Class 10: Quadratic Equations – CBSE Board Problems

$\displaystyle \textbf{Question 1. }\text{The sides of a right triangle are such that the longest side is }4\text{ m more than}$
$\displaystyle \text{the shortest side and the third side is }2\text{ m less than the longest side.}$
$\displaystyle \text{Find the length of each side of the triangle. Also, find the difference between the}$
$\displaystyle \text{numerical values of the area and the perimeter of the given triangle.} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }ABC\text{ is a right angled triangle.}$
$\displaystyle \text{Let shortest side }BC=x\text{ m}$
$\displaystyle AC=(x+4)\text{ m}$
$\displaystyle AB=(x+4)-2=(x+2)\text{ m}$
$\displaystyle \text{To find : Length of each side of the triangle}$
$\displaystyle \text{As }ABC\text{ is a right angled triangle. So it follows Pythagoras theorem.}$
$\displaystyle AC^{2}=AB^{2}+BC^{2}$
$\displaystyle (x+4)^{2}=(x+2)^{2}+x^{2}$
$\displaystyle x^{2}+16+8x=x^{2}+4+4x+x^{2}$
$\displaystyle \Rightarrow x^{2}-4x-12=0$
$\displaystyle \Rightarrow x^{2}-6x+2x-12=0$
$\displaystyle \Rightarrow (x+2)(x-6)=0$
$\displaystyle \Rightarrow x=-2,\ x=6$
$\displaystyle x=-2,\ \text{not possible}$
$\displaystyle \therefore \text{Shortest side}=6\text{ m}$
$\displaystyle \text{Longest side}=6+4=10\text{ m}$
$\displaystyle \text{Third side}=10-2=8\text{ m}$
$\displaystyle \text{Also area of }\triangle ABC=\frac{1}{2}\times b\times h=\frac{1}{2}\times6\times8$
$\displaystyle =24\text{ sq units}$
$\displaystyle \text{Perimeter of }\triangle ABC=10+8+6=24$
$\displaystyle \text{Now area of }\triangle ABC-\text{Perimeter of }\triangle ABC=24-24=0$
$\\$

$\displaystyle \textbf{Question 2. }\text{Express the equation }\frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3};\ (x\neq 3,5)\text{ as a quadratic}$
$\displaystyle \text{equation in standard form. Hence, find the roots of the equation so formed.} \\ \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is}$
$\displaystyle \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3},\ (x\neq3,5)$
$\displaystyle \text{We have to change it into standard form of equation}$
$\displaystyle \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3}$
$\displaystyle \Rightarrow \frac{x^{2}-5x-2x+10+x^{2}-3x-4x+12}{x^{2}-5x-3x+15}=\frac{10}{3}$
$\displaystyle \Rightarrow \frac{2x^{2}-14x+22}{x^{2}-8x+15}=\frac{10}{3}$
$\displaystyle \Rightarrow 6x^{2}-42x+66=10x^{2}-80x+150$
$\displaystyle \Rightarrow 4x^{2}-38x+84=0$
$\displaystyle \Rightarrow 2x^{2}-19x+42=0$
$\displaystyle \Rightarrow 2x^{2}-12x-7x+42=0$
$\displaystyle \Rightarrow 2x(x-6)-7(x-6)=0$
$\displaystyle \Rightarrow (2x-7)(x-6)=0$
$\displaystyle \therefore x=\frac{7}{2},\ x=6$
$\\$

$\displaystyle \textbf{Question 3. }\text{The value of }a\text{ for which }ax^{2}+x+a=0 \\ \text{ has equal and positive roots is:} \hspace{0.2cm}\text{[CBSE 2025]}$
$\displaystyle \text{(a) }2 \qquad \text{(b) }-2 \qquad \text{(c) }\frac{1}{2} \qquad \text{(d) }-\frac{1}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) The given quadratic equation is,}$
$\displaystyle ax^{2}+x+a=0$
$\displaystyle \text{For equal and positive roots, }D=0$
$\displaystyle \text{i.e. }b^{2}-4ac=0$
$\displaystyle (1)^{2}-4(a)(a)=0$
$\displaystyle \Rightarrow 4a^{2}=1$
$\displaystyle \Rightarrow a^{2}=\frac{1}{4}$
$\displaystyle \Rightarrow a=\pm\frac{1}{2}$
$\displaystyle \Rightarrow a=-\frac{1}{2}\text{ for positive roots}$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }x=5\text{ is a solution of the quadratic equation } \\ 2x^{2}+(k-1)x+10=0,\text{ then the value of }k\text{ is:} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }11 \qquad \text{(b) }-11 \qquad \text{(c) }13 \qquad \text{(d) }-13$
$\displaystyle \text{Answer:}$
$\displaystyle (b)\ \text{Since }x=5\text{ is a solution of }2x^{2}+(k-1)x+10=0$
$\displaystyle \Rightarrow 50+5(k-1)+10=0$
$\displaystyle \Rightarrow 5(k-1)=-60$
$\displaystyle \Rightarrow k=-11$
$\\$

$\displaystyle \textbf{Question 5. }\text{If the roots of equation }ax^{2}+bx+c=0,\ a\neq0 \\ \text{ are real and equal, then which of the following}$
$\displaystyle \text{relation is true?} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(a) }a=\frac{b^{2}}{c} \qquad \text{(b) }b^{2}=ac$
$\displaystyle \text{(c) }ac=\frac{b^{2}}{4} \qquad \text{(d) }c=\frac{b^{2}}{a}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) The given quadratic equation is,}$
$\displaystyle ax^{2}+bx+c=0$
$\displaystyle \text{For real and equal roots,}$
$\displaystyle \text{Discriminant, }D=0\Rightarrow b^{2}-4ac=0$
$\displaystyle \Rightarrow ac=\frac{b^{2}}{4}$
$\\$

$\displaystyle \textbf{Question 6. }\text{A rectangular floor area can be completely tiled with }200\text{ square tiles. If}$
$\displaystyle \text{the side length of each tile is increased by }1\text{ unit, it would take only }128\text{ tiles to cover}$
$\displaystyle \text{the floor.} \hspace{0.2cm}\text{[CBSE 2024]}$
$\displaystyle \text{(i) Assuming the original length of each side of a tile be }x\text{ units, make a quadratic}$
$\displaystyle \text{equation from the above information.}$
$\displaystyle \text{(ii) Write the corresponding quadratic equation in standard form.}$
$\displaystyle \text{(iii) (a) Find the value of }x,\text{ the length of side of a tile by factorisation.}$
$\displaystyle \text{OR}$
$\displaystyle \text{(iii) (b) Solve the quadratic equation for }x,\text{ using quadratic formula.}$
$\displaystyle \text{Answer:}$
$\displaystyle \textbf{(i) }\text{Original length of each side of a square tile}=x\text{ units}$
$\displaystyle \text{So, original area of each square tile}=x^{2}\text{ sq units}$
$\displaystyle \text{Now, increased length of each side of a square tile}=(x+1)\text{ units}$
$\displaystyle \therefore \text{New area of each square tile}=(x+1)^{2}\text{ sq units}$
$\displaystyle \text{ATQ, }200\times x^{2}=128(x+1)^{2}$

$\displaystyle \textbf{(ii) }\text{We have,}$
$\displaystyle 200x^{2}=128(x+1)^{2}$
$\displaystyle \Rightarrow 200x^{2}=128(x^{2}+1+2x)$
$\displaystyle \Rightarrow 200x^{2}=128x^{2}+128+256x$
$\displaystyle \Rightarrow 72x^{2}-256x-128=0$
$\displaystyle \Rightarrow 8(9x^{2}-32x-16)=0$
$\displaystyle \Rightarrow 9x^{2}-32x-16=0$
$\displaystyle \text{This is the required quadratic equation in standard form.}$

$\displaystyle \textbf{(iii)(a) }\text{We have, }9x^{2}-32x-16=0$
$\displaystyle \Rightarrow 9x^{2}-36x+4x-16=0$
$\displaystyle \Rightarrow 9x(x-4)+4(x-4)=0$
$\displaystyle \Rightarrow (x-4)(9x+4)=0$
$\displaystyle \Rightarrow x-4=0\text{ or }9x+4=0$
$\displaystyle \Rightarrow x=4\text{ or }x=-\frac{4}{9}$
$\displaystyle \left(\text{Rejected as side cannot be negative}\right)$
$\displaystyle \therefore \text{Length of each side of a square tile}=4\text{ units.}$

$\displaystyle \textbf{(iii)(b) }\text{We have, }9x^{2}-32x-16=0$
$\displaystyle \text{Comparing with }ax^{2}+bx+c=0,\text{ we get}$
$\displaystyle a=9,\ b=-32,\ c=-16$
$\displaystyle \text{Now, discriminant,}$
$\displaystyle D=b^{2}-4ac$
$\displaystyle =(-32)^{2}-4\times9\times(-16)$
$\displaystyle =1024+576=1600$
$\displaystyle \text{By quadratic formula,}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle \Rightarrow x=\frac{32\pm\sqrt{1600}}{2\times9}$
$\displaystyle \Rightarrow x=\frac{32+40}{18}\text{ or }x=\frac{32-40}{18}$
$\displaystyle \Rightarrow x=4\text{ or }x=-\frac{4}{9}$
$\\$

$\displaystyle \textbf{Question 7. }\text{The roots of the equation }x^{2}+3x-10=0\text{ are:} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{(a) }2,-5 \qquad \text{(b) }-2,5 \qquad \text{(c) }2,5 \qquad \text{(d) }-2,-5$
$\displaystyle \text{Answer:}$
$\displaystyle (a)\ x^{2}+3x-10=0$
$\displaystyle \Rightarrow (x+5)(x-2)=0$
$\displaystyle \therefore x=-5,\ 2$
$\displaystyle \therefore \text{Roots are }-5\text{ and }2.$
$\\$

$\displaystyle \textbf{Question 8. }\text{Three consecutive odd numbers are such that the sum of the squares}$
$\displaystyle \text{of the first two numbers is greater than the square of the third by }65.\text{ Find the numbers.} \\ \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the three consecutive numbers be }(2x-1),\ (2x+1)\text{ and }(2x+3).$
$\displaystyle \text{ATQ, }(2x-1)^{2}+(2x+1)^{2}=(2x+3)^{2}+65$
$\displaystyle \Rightarrow 8x^{2}+2=4x^{2}+9+12x+65$
$\displaystyle \Rightarrow 4x^{2}-12x-72=0$
$\displaystyle \Rightarrow x^{2}-3x-18=0$
$\displaystyle \Rightarrow x^{2}-6x+3x-18=0$
$\displaystyle \Rightarrow x(x-6)+3(x-6)=0$
$\displaystyle \Rightarrow (x-6)(x+3)=0$
$\displaystyle \therefore x=6\ \text{or}\ x=-3$
$\displaystyle \text{For }x=6,\ \text{the numbers are }11,\ 13\text{ and }15.$
$\displaystyle \text{For }x=-3,\ \text{the numbers are }-7,\ -5\text{ and }-3.$
$\\$

$\displaystyle \textbf{Question 9. }\text{The values of }k\text{ for which the equation } \\ 2x^{2}-kx+1=0\text{ has real and equal roots is/are:} \hspace{0.2cm}\text{[CBSE 2023(C)]}$
$\displaystyle \text{(a) }2\sqrt{2} \qquad \text{(b) }-2\sqrt{2}$
$\displaystyle \text{(c) }\pm2\sqrt{2} \qquad \text{(d) }2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is }2x^{2}-kx+1=0$
$\displaystyle \text{We get, }a=2,\ b=-k,\ c=1$
$\displaystyle D=b^{2}-4ac=(-k)^{2}-4\times2\times1$
$\displaystyle =k^{2}-8$
$\displaystyle \text{For real and equal roots,}$
$\displaystyle D=0$
$\displaystyle \Rightarrow k^{2}-8=0$
$\displaystyle \Rightarrow k=\pm2\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 10. }\text{Find the sum and product of the roots of the quadratic equation } \\ 2x^{2}-9x+4=0. \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given quadratic equation is, }2x^{2}-9x+4=0$
$\displaystyle \text{Here, }a=2,\ b=-9,\ c=4$
$\displaystyle \text{Now, sum of roots}=-\frac{b}{a}=-\frac{-9}{2}=\frac{9}{2}$
$\displaystyle \text{and}$
$\displaystyle \text{Product of roots}=\frac{c}{a}=\frac{4}{2}=2$
$\\$

$\displaystyle \textbf{Question 11. }\text{Find the discriminant of the quadratic equation }4x^{2}-5=0$
$\displaystyle \text{ and hence comment on the nature of }\text{roots of the equation.} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }4x^{2}-5=0$
$\displaystyle \Rightarrow a=4,\ b=0,\ c=-5$
$\displaystyle D=b^{2}-4ac$
$\displaystyle =0-4\times4\times(-5)=80>0$
$\displaystyle \therefore \text{Roots are real and distinct.}$
$\\$

$\displaystyle \textbf{Question 12. }\text{Find the value of }p\text{ for which the quadratic}$
$\displaystyle \text{equation }px(x-2)+6=0\text{ has two equal real roots.} \hspace{0.2cm}\text{[CBSE 2023]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{p}{x(x-2)}+6=0$
$\displaystyle \Rightarrow px^{2}-2px+6=0$
$\displaystyle \text{Here, }a=p,\ b=-2p,\ c=6$
$\displaystyle D=b^{2}-4ac=4p^{2}-24p$
$\displaystyle \text{For equal roots,}$
$\displaystyle D=0$
$\displaystyle \Rightarrow 4p^{2}-24p=0$
$\displaystyle \Rightarrow 4p(p-6)=0$
$\displaystyle \Rightarrow p=0\text{ or }p=6$
$\displaystyle \text{But }p\neq0\text{ (In a quadratic equation }ax^{2}+bx+c=0,\ a\neq0).$
$\displaystyle \therefore p=6$
$\\$

$\displaystyle \textbf{Question 13. }\text{Solve the quadratic equation: }x^{2}+2\sqrt{2}x-6=0\text{ for }x. \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \ x^{2}+2\sqrt{2}x-6=0$
$\displaystyle =x^{2}+3\sqrt{2}x-\sqrt{2}x-6$
$\displaystyle =x(x+3\sqrt{2})-\sqrt{2}(x+3\sqrt{2})$
$\displaystyle =(x+3\sqrt{2})(x-\sqrt{2})$
$\displaystyle \Rightarrow x+3\sqrt{2}=0\ \text{or}\ x-\sqrt{2}=0$
$\displaystyle \therefore x=-3\sqrt{2}\ \text{or}\ x=\sqrt{2}$
$\\$

$\displaystyle \textbf{Question 14. }\text{A }2\text{-digit number is such that the product of its digits is 24. If 18 is }$
$\displaystyle \text{subtracted from the number, the digits interchange their places. Find the number.} \\ \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one's digit be }x\text{ and ten's digit be }y.$
$\displaystyle \therefore \text{Number}=10y+x$
$\displaystyle \text{According to the question}$
$\displaystyle xy=24 \qquad (i)$
$\displaystyle \Rightarrow 10y+x-18=10x+y$
$\displaystyle \Rightarrow 9y-9x=18$
$\displaystyle \Rightarrow y-x=2$
$\displaystyle \Rightarrow y=2+x \qquad (ii)$
$\displaystyle \text{Putting the value of }y\text{ in }(i),\ \text{we get}$
$\displaystyle x(2+x)=24$
$\displaystyle \Rightarrow x^{2}+2x-24=0$
$\displaystyle \Rightarrow (x+6)(x-4)=0$
$\displaystyle \Rightarrow x=4\text{ or }x=-6$
$\displaystyle \text{If }x=4,\ y=6,\ \text{Number}=10\times6+4=64$
$\displaystyle \text{If }x=-6,\ y=-4,\ \text{Number}=10\times(-4)-6=-46$
$\\$

$\displaystyle \textbf{Question 15. }\text{The difference of the squares of two numbers is 180. The square of the }$
$\displaystyle \text{smaller number is }8\text{ times the greater number. } \text{Find the two numbers.} \hspace{0.2cm}\text{[CBSE 2022]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the two numbers be }x\text{ and }y.$
$\displaystyle x^{2}-y^{2}=180\ (\text{let }x>y) \qquad (i)$
$\displaystyle y^{2}=8x \qquad (ii)$
$\displaystyle \text{Putting the value in }(i),\ \text{we get}$
$\displaystyle x^{2}-8x-180=0$
$\displaystyle \Rightarrow x^{2}-18x+10x-180=0$
$\displaystyle \Rightarrow x(x-18)+10(x-18)=0$
$\displaystyle \Rightarrow (x-18)(x+10)=0$
$\displaystyle \Rightarrow x=18\text{ or }x=-10$
$\displaystyle \text{If }x=18,\ \text{then }y^{2}=18\times8 \Rightarrow y=\pm12$
$\displaystyle \text{If }x=-10,\ \text{then }y^{2}=-10\times8=-80\ (\text{impossible})$
$\displaystyle \therefore \text{The numbers are }18\text{ and }12\text{ or }18\text{ and }-12.$
$\\$

$\displaystyle \textbf{Question 16. }\text{In a flight of }600\text{ km, an aircraft was slowed due to bad weather.}$
$\displaystyle \text{Its average speed for the trip was reduced by }200\text{ km/hr} \text{ and time of flight increased by }$
$\displaystyle 30\text{ minutes. } \text{Find the original duration of flight.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let original speed of the aircraft be }x\text{ km/hr.}$
$\displaystyle \text{Reduced speed}=(x-200)\text{ km/hr}$
$\displaystyle \text{According to given condition,}$
$\displaystyle \frac{600}{x-200}-\frac{600}{x}=\frac{30}{60}=\frac{1}{2}$
$\displaystyle \frac{600x-600x+120000}{x(x-200)}=\frac{1}{2}$
$\displaystyle \Rightarrow x^{2}-200x-240000=0$
$\displaystyle \Rightarrow (x+400)(x-600)=0$
$\displaystyle \Rightarrow x+400=0\text{ or }x-600=0$
$\displaystyle \Rightarrow x=-400\text{ (rejected) or }x=600$
$\displaystyle \therefore \text{Original speed}=600\text{ km/hr}$
$\displaystyle \therefore \text{Original duration of flight}=\frac{600}{600}=1\text{ hour}$
$\\$

$\displaystyle \textbf{Question 17. }\text{A train covers a distance of }480\text{ km at a uniform speed. If the speed }$
$\displaystyle \text{had been }8\text{ km/hr less, } \text{then it would have taken }3\text{ hours more}$
$\displaystyle \text{to cover the same distance. } \text{Find the original speed of the train.} \hspace{0.2cm}\text{[CBSE 2020]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let original speed of train be }x\text{ km/hr.}$
$\displaystyle \text{Now, new speed be }(x-8)\text{ km/hr}$
$\displaystyle \therefore \frac{480}{x-8}-\frac{480}{x}=3$
$\displaystyle \frac{480x-480x+3840}{x(x-8)}=3$
$\displaystyle \Rightarrow 3x(x-8)=3840$
$\displaystyle \Rightarrow x^{2}-8x-1280=0$
$\displaystyle \Rightarrow x^{2}-40x+32x-1280=0$
$\displaystyle \Rightarrow x(x-40)+32(x-40)=0$
$\displaystyle \Rightarrow (x+32)(x-40)=0$
$\displaystyle \Rightarrow x+32=0\text{ or }x-40=0$
$\displaystyle \Rightarrow x=-32\text{ (rejected) or }x=40$
$\displaystyle \therefore \text{Original speed of train}=40\text{ km/hr}$
$\\$

$\displaystyle \textbf{Question 18. }\text{Two water taps together can fill a tank in }1\frac{7}{8}\text{ hours. The tap with longer}$
$\displaystyle \text{diameter takes }2\text{ hours less than the tap with smaller one to fill the tank separately.}$
$\displaystyle \text{Find the time in which each tap can fill the tank separately.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let time taken by tap with longer diameter}=x\text{ hours}$
$\displaystyle \text{Time taken by tap with smaller diameter}=(x+2)\text{ hours}$
$\displaystyle \text{As per condition,}$
$\displaystyle \frac{1}{x}+\frac{1}{x+2}=\frac{8}{15}$
$\displaystyle \Rightarrow \frac{x+2+x}{x(x+2)}=\frac{8}{15}$
$\displaystyle \Rightarrow \frac{2x+2}{x(x+2)}=\frac{8}{15}$
$\displaystyle \Rightarrow 15(x+1)=4x(x+2)$
$\displaystyle \Rightarrow 15(x+1)=4x^{2}+8x$
$\displaystyle \Rightarrow 4x^{2}-7x-15=0$
$\displaystyle \Rightarrow 4x^{2}-12x+5x-15=0$
$\displaystyle \Rightarrow 4x(x-3)+5(x-3)=0$
$\displaystyle \Rightarrow (x-3)(4x+5)=0$
$\displaystyle \Rightarrow x=3,\ x=-\frac{5}{4}\text{ (not possible)}$
$\displaystyle \text{Tap with longer diameter fills the tank in }3\text{ hours.}$
$\displaystyle \text{Tap with smaller diameter fills the tank in }5\text{ hours.}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Solve the following for }x: \ \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{2a+b+2x}=\frac{1}{2a}+\frac{1}{b}+\frac{1}{2x}$
$\displaystyle \Rightarrow \frac{1}{2a+b+2x}-\frac{1}{2x}=\frac{1}{2a}+\frac{1}{b}$
$\displaystyle \Rightarrow \frac{2x-2a-b-2x}{(2a+b+2x)(2x)}=\frac{b+2a}{2ab}$
$\displaystyle \Rightarrow \frac{-(2a+b)}{(2a+b+2x)2x}=\frac{b+2a}{2ab}$
$\displaystyle \Rightarrow \frac{-1}{4ax+2bx+4x^{2}}=\frac{1}{2ab}$
$\displaystyle \Rightarrow 4x^{2}+2bx+4ax+2ab=0$
$\displaystyle \Rightarrow 2x(2x+b)+2a(2x+b)=0$
$\displaystyle \Rightarrow (2x+b)(2x+2a)=0$
$\displaystyle \Rightarrow x=-\frac{b}{2}\text{ or }x=-a$
$\\$

$\displaystyle \textbf{Question 20. }\text{For what values of }k,\text{ the roots of the equation}$
$\displaystyle x^{2}+4x+k=0\text{ are real?} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is }x^{2}+4x+k=0$
$\displaystyle D=b^{2}-4ac$
$\displaystyle D=(4)^{2}-4\times1\times k=16-4k$
$\displaystyle \text{For real roots }D\geq0$
$\displaystyle \Rightarrow 16-4k\geq0$
$\displaystyle \Rightarrow 16\geq4k$
$\displaystyle \Rightarrow k\leq4$
$\\$

$\displaystyle \textbf{Question 21. }\text{Find the value of }k\text{ for which the roots of the equation}$
$\displaystyle 3x^{2}-10x+k=0\text{ are reciprocal of each other.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given equation is }3x^{2}-10x+k=0$
$\displaystyle \text{Let one root of the equation is }\alpha\text{ and another root is }\frac{1}{\alpha}$
$\displaystyle \text{Product of the roots}=\alpha\cdot\frac{1}{\alpha}$
$\displaystyle \text{So, }\alpha\cdot\frac{1}{\alpha}=\frac{k}{3}$
$\displaystyle \Rightarrow 1=\frac{k}{3}$
$\displaystyle \Rightarrow k=3$
$\\$

$\displaystyle \textbf{Question 22. }\text{Write all the values of }p\text{ for which the } \text{equation }$
$\displaystyle x^{2}+px+16=0 \text{ has equal roots. Find } \text{the roots of the equation so obtained.} \hspace{0.2cm}\text{[CBSE 2019]}$
$\displaystyle \text{Answer:}$
$\displaystyle x^{2}+px+16=0$
$\displaystyle \therefore D=b^{2}-4ac$
$\displaystyle D=p^{2}-4\times1\times16$
$\displaystyle D=p^{2}-64$
$\displaystyle \text{For equal roots,}$
$\displaystyle D=0$
$\displaystyle \Rightarrow p^{2}-64=0$
$\displaystyle \Rightarrow p=\pm8$
$\displaystyle \text{For }p=8,\ \text{we have}$
$\displaystyle x^{2}+8x+16=0\Rightarrow (x+4)^{2}=0$
$\displaystyle \Rightarrow x=-4,-4$
$\displaystyle \text{For }p=-8,\ \text{we have}$
$\displaystyle x^{2}-8x+16=0\Rightarrow (x-4)^{2}=0$
$\displaystyle \Rightarrow x=4,4$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }x=3\text{ is one root of the quadratic equation } \\ x^{2}-2kx-6=0,\text{ then find the value of }k. \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Putting }x=3\text{ in }x^{2}-2kx-6=0,\ \text{we get}$
$\displaystyle 3^{2}-2k\times3-6=0$
$\displaystyle \Rightarrow 9-6k-6=0$
$\displaystyle \Rightarrow 3=6k$
$\displaystyle \therefore k=\frac{3}{6}=\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 24. }\text{A plane left }30\text{ minutes later than its scheduled time. In order to reach }$
$\displaystyle \text{the destination }1500\text{ km away in time, } \text{it had to increase its speed by }100\text{ km/h}$
$\displaystyle \text{from the usual speed. } \text{Find its usual speed.} \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the usual speed be }x\text{ km/h and distance is }1500\text{ km.}$
$\displaystyle \text{Now, }30\text{ minutes}=\frac{30}{60}\text{ hours}=\frac{1}{2}\text{ hours}$
$\displaystyle \text{Then time taken}=\frac{\text{Distance travelled}}{\text{Speed}}$
$\displaystyle =\frac{1500}{x}\text{ hours}$
$\displaystyle \text{New increased speed}=(x+100)\text{ km/h}$
$\displaystyle \text{Then new time taken}=\frac{1500}{x+100}\text{ hours}$
$\displaystyle \text{ATQ, }\frac{1500}{x}-\frac{1500}{x+100}=\frac{1}{2}$
$\displaystyle \Rightarrow \frac{1500[x+100-x]}{x(x+100)}=\frac{1}{2}$
$\displaystyle \Rightarrow x^{2}+100x-300000=0$
$\displaystyle \Rightarrow (x+600)(x-500)=0$
$\displaystyle \Rightarrow x+600=0\text{ or }x-500=0$
$\displaystyle \Rightarrow x=-600\text{ (rejected) or }x=500$
$\displaystyle \therefore \text{Usual speed of the aeroplane}=500\text{ km/h}$
$\\$

$\displaystyle \textbf{Question 25. }\text{Divide }27\text{ into two parts such that the sum of their reciprocals is } \\ \frac{3}{20}. \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let one part be }x\text{ and another part be }27-x.$
$\displaystyle \text{So, }\frac{1}{x}+\frac{1}{27-x}=\frac{3}{20}$
$\displaystyle \Rightarrow \frac{27-x+x}{x(27-x)}=\frac{3}{20}$
$\displaystyle \Rightarrow \frac{27}{x(27-x)}=\frac{3}{20}$
$\displaystyle \Rightarrow 180=x(27-x)$
$\displaystyle \Rightarrow x^{2}-27x+180=0$
$\displaystyle \Rightarrow (x-15)(x-12)=0$
$\displaystyle \Rightarrow x=15,\ x=12$
$\displaystyle \text{So, two parts are }15\text{ and }12.$
$\\$

$\displaystyle \textbf{Question 26. }\text{A motor boat whose speed is }18\text{ km/hr in still water takes }1\text{ hr more to go }$
$\displaystyle 24\text{ km upstream than to return downstream to the same spot. } \text{Find the speed of the}$
$\displaystyle \text{stream.} \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let speed of stream be }x\text{ km/h.}$
$\displaystyle \text{Speed of motor boat in still water}=18\text{ km/h}$
$\displaystyle \text{Upstream speed}=(18-x)\text{ km/h}$
$\displaystyle \text{Downstream speed}=(18+x)\text{ km/h}$
$\displaystyle \text{Distance}=24\text{ km}$
$\displaystyle \text{Time taken for upstream}=\frac{24}{18-x}\text{ hours}$
$\displaystyle \text{Time taken for downstream}=\frac{24}{18+x}\text{ hours}$
$\displaystyle \text{ATQ, }\frac{24}{18-x}-\frac{24}{18+x}=1$
$\displaystyle \Rightarrow \frac{24[(18+x-18+x)]}{(18-x)(18+x)}=1$
$\displaystyle \Rightarrow \frac{24\times2x}{18^{2}-x^{2}}=1$
$\displaystyle \Rightarrow x^{2}+48x-324=0$
$\displaystyle \Rightarrow x^{2}+54x-6x-324=0$
$\displaystyle \Rightarrow x(x+54)-6(x+54)=0$
$\displaystyle \Rightarrow (x+54)(x-6)=0$
$\displaystyle \Rightarrow x=-54\text{ (rejected) or }x=6$
$\displaystyle \therefore \text{Speed of stream}=6\text{ km/h}$
$\\$

$\displaystyle \textbf{Question 27. }\text{A train travels at a certain average speed for a distance of }63\text{ km and then}$
$\displaystyle \text{travels a distance of }72\text{ km at an average speed of }6\text{ km/hr more than its original}$
$\displaystyle \text{speed. If it takes }3\text{ hours to complete total journey, what is the original average}$
$\displaystyle \text{speed?} \hspace{0.2cm}\text{[CBSE 2018]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let original average speed of the train be }x\text{ km/h.}$
$\displaystyle \text{New average speed}=(x+6)\text{ km/h}$
$\displaystyle \text{Time taken for a distance of }63\text{ km}=\frac{63}{x}\text{ hours}$
$\displaystyle \text{Time taken for a distance of }72\text{ km}=\frac{72}{x+6}\text{ hours}$
$\displaystyle \text{ATQ, }\frac{63}{x}+\frac{72}{x+6}=3$
$\displaystyle \Rightarrow \frac{63(x+6)+72x}{x(x+6)}=3$
$\displaystyle \Rightarrow \frac{63x+378+72x}{x^{2}+6x}=3$
$\displaystyle \Rightarrow 3(x^{2}+6x)=135x+378$
$\displaystyle \Rightarrow x^{2}+6x=45x+126$
$\displaystyle \Rightarrow x^{2}-39x-126=0$
$\displaystyle \Rightarrow x^{2}-42x+3x-126=0$
$\displaystyle \Rightarrow x(x-42)+3(x-42)=0$
$\displaystyle \Rightarrow (x-42)(x+3)=0$
$\displaystyle \Rightarrow x=42\text{ or }x=-3\text{ (rejected)}$
$\displaystyle \therefore \text{Original average speed of the train}=42\text{ km/h}$
$\\$

$\displaystyle \textbf{Question 28. }\text{A faster train takes one hour less than a slower train for a journey of }$
$\displaystyle 200\text{ km. }\text{If the speed of slower train is }10\text{ km/hr less than that of faster train, find the}$
$\displaystyle \text{speed of two trains.} \hspace{0.2cm}\text{[CBSE 2018(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let speed of faster train be }x\text{ km/hr.}$
$\displaystyle \text{So, speed of slower train}=(x-10)\text{ km/hr}$
$\displaystyle \text{Distance}=200\text{ km}$
$\displaystyle \text{Time taken by slower train}=\left(\frac{200}{x-10}\right)\text{ hrs}$
$\displaystyle \text{Time taken by faster train}=\left(\frac{200}{x}\right)\text{ hrs}$
$\displaystyle \text{As per condition,}$
$\displaystyle \frac{200}{x-10}-\frac{200}{x}=1$
$\displaystyle \Rightarrow 200\left[\frac{x-x+10}{x^{2}-10x}\right]=1$
$\displaystyle \Rightarrow x^{2}-10x-2000=0$
$\displaystyle \Rightarrow x^{2}-50x+40x-2000=0$
$\displaystyle \Rightarrow (x-50)(x+40)=0$
$\displaystyle \Rightarrow x=50\text{ and }x=-40\text{ (not possible)}$
$\displaystyle \therefore \text{Speed of faster train}=50\text{ km/hr}$
$\displaystyle \therefore \text{Speed of slower train}=40\text{ km/hr}$
$\\$

$\displaystyle \textbf{Question 29. }\text{Speed of a boat in still water is }15\text{ km/h. It goes }30\text{ km upstream and}$
$\displaystyle \text{returns back at the same point in }4\text{ hours }30\text{ minutes. Find the speed of the}$
$\displaystyle \text{stream.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Speed of boat in still water}=15\text{ km/h}$
$\displaystyle \text{Let speed of the stream be }x\text{ km/h.}$
$\displaystyle \text{Then speed of the boat in downstream}=(15+x)\text{ km/h}$
$\displaystyle \text{and speed of the boat in upstream}=(15-x)\text{ km/h}$
$\displaystyle \text{Distance}=30\text{ km}$
$\displaystyle \text{ATQ, }\frac{30}{15-x}+\frac{30}{15+x}=\frac{4\times30}{60}$
$\displaystyle \Rightarrow 30\left[\frac{1}{15-x}+\frac{1}{15+x}\right]=\frac{270}{60}=\frac{27}{6}$
$\displaystyle \Rightarrow \frac{1}{15-x}+\frac{1}{15+x}=\frac{27}{6\times30}$
$\displaystyle \Rightarrow \frac{1}{15-x}+\frac{1}{15+x}=\frac{3}{20}$
$\displaystyle \Rightarrow \frac{15+x+15-x}{(15-x)(15+x)}=\frac{3}{20}$
$\displaystyle \Rightarrow \frac{30}{225-x^{2}}=\frac{3}{20}$
$\displaystyle \Rightarrow 225-x^{2}=200$
$\displaystyle \Rightarrow x^{2}=25$
$\displaystyle \Rightarrow x=5\text{ or }-5$
$\displaystyle x=-5\text{ is rejected because speed of stream cannot be negative}$
$\displaystyle \therefore x=5\text{ km/h}$
$\displaystyle \text{Hence, speed of the stream is }5\text{ km/h}.$
$\\$

$\displaystyle \textbf{Question 30. }\text{A train covers a distance of }300\text{ km at a uniform speed. If the speed of}$
$\displaystyle \text{the train is increased by }5\text{ km/hour, it takes }2\text{ hours less in the journey. Find the }$
$\displaystyle \text{original speed of the train.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the train be }x\text{ km/hr.}$
$\displaystyle \text{It has to cover a distance of }300\text{ km.}$
$\displaystyle \therefore \text{Time taken to cover }300\text{ km}=\frac{300}{x}\text{ hrs}$
$\displaystyle \text{When the speed is increased by }5\text{ km/hr, the speed becomes }(x+5)\text{ km/hr}$
$\displaystyle \therefore \text{Time taken to cover }300\text{ km at the new speed}=\frac{300}{x+5}\text{ hrs}$
$\displaystyle \text{It is given that the train reaches }2\text{ hrs early.}$
$\displaystyle \therefore \frac{300}{x}-\frac{300}{x+5}=2$
$\displaystyle \Rightarrow 300\left[\frac{1}{x}-\frac{1}{x+5}\right]=2$
$\displaystyle \Rightarrow \frac{x+5-x}{x(x+5)}=\frac{1}{150}$
$\displaystyle \Rightarrow \frac{5}{x^{2}+5x}=\frac{1}{150}$
$\displaystyle \Rightarrow x^{2}+5x-750=0$
$\displaystyle \Rightarrow x^{2}+30x-25x-750=0$
$\displaystyle \Rightarrow x(x+30)-25(x+30)=0$
$\displaystyle \Rightarrow (x-25)(x+30)=0$
$\displaystyle \Rightarrow x=25\text{ or }x=-30$
$\displaystyle \text{But speed cannot be negative.}$
$\displaystyle \therefore x=25$
$\displaystyle \text{Hence, the original speed of the train is }25\text{ km/hr}.$
$\\$

$\displaystyle \textbf{Question 31. }\text{Solve for }x: \frac{x-1}{2x+1}+\frac{2x+1}{x-1}=2,\text{ where }x\neq-\frac{1}{2},1 \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-1}{2x+1}+\frac{2x+1}{x-1}=2 \qquad (i)$
$\displaystyle \text{Put }\frac{x-1}{2x+1}=y$
$\displaystyle \therefore \frac{2x+1}{x-1}=\frac{1}{y}$
$\displaystyle \text{Equation }(i)\text{ becomes}$
$\displaystyle y+\frac{1}{y}=2$
$\displaystyle \Rightarrow \frac{y^{2}+1}{y}=2$
$\displaystyle \Rightarrow y^{2}-2y+1=0$
$\displaystyle \Rightarrow y^{2}-y-y+1=0$
$\displaystyle \Rightarrow (y-1)(y-1)=0$
$\displaystyle \Rightarrow y=1,\ 1$
$\displaystyle \text{Bringing back the value of }y,\ \text{we get}$
$\displaystyle \frac{x-1}{2x+1}=1\ \text{and}\ \frac{2x+1}{x-1}=1$
$\displaystyle \Rightarrow x-1=2x+1\ \text{and}\ 2x+1=x-1$
$\displaystyle \Rightarrow x=-2,\ -2$
$\\$

$\displaystyle \textbf{Question 32. }\text{A takes }6\text{ days less than B to do a work. If both A and B working}$
$\displaystyle \text{together can do it in }4\text{ days, how } \text{many days will B take to finish it?} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of days taken by }B\text{ alone to finish the work be }x.$
$\displaystyle \text{Then }A\text{ alone will take }(x-6)\text{ days to finish the same work.}$
$\displaystyle \therefore \text{Portion of work done by }B\text{ alone in one day}=\frac{1}{x}$
$\displaystyle \text{and portion of work done by }A\text{ alone in one day}=\frac{1}{x-6}$
$\displaystyle \text{If }A\text{ and }B\text{ work together, then they take four days to finish the work.}$
$\displaystyle \therefore \text{Portion of work done by }A\text{ and }B\text{ together in one day}=\frac{1}{4}$
$\displaystyle \therefore \frac{1}{x}+\frac{1}{x-6}=\frac{1}{4}$
$\displaystyle \Rightarrow \frac{(x-6)+x}{x(x-6)}=\frac{1}{4}$
$\displaystyle \Rightarrow \frac{2x-6}{x^{2}-6x}=\frac{1}{4}$
$\displaystyle \Rightarrow x^{2}-14x+24=0$
$\displaystyle \Rightarrow x^{2}-12x-2x+24=0$
$\displaystyle \Rightarrow x(x-12)-2(x-12)=0$
$\displaystyle \Rightarrow (x-2)(x-12)=0$
$\displaystyle \Rightarrow x=2\text{ or }x=12$
$\displaystyle x=2\text{ is rejected because }x-6\text{ will give a negative number of days, which is not possible.}$
$\displaystyle \text{Hence, }x=12$
$\displaystyle \text{Thus, }B\text{ takes }12\text{ days and }A\text{ takes }x-6\text{ i.e. }12-6=6\text{ days to finish the work individually.}$
$\\$

$\displaystyle \textbf{Question 33. }\text{Solve for }x: \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x},\ x\neq0,1,2 \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{x-2}+\frac{2}{x-1}=\frac{6}{x}$
$\displaystyle \Rightarrow \frac{x-1+2x-4}{(x-2)(x-1)}=\frac{6}{x}$
$\displaystyle \Rightarrow \frac{3x-5}{x^{2}-3x+2}=\frac{6}{x}$
$\displaystyle \Rightarrow x(3x-5)=6(x^{2}-3x+2)$
$\displaystyle \Rightarrow 3x^{2}-13x+12=0$
$\displaystyle \Rightarrow 3x^{2}-9x-4x+12=0$
$\displaystyle \Rightarrow 3x(x-3)-4(x-3)=0$
$\displaystyle \Rightarrow (x-3)(3x-4)=0$
$\displaystyle \Rightarrow x=3\text{ or }x=\frac{4}{3}$
$\\$

$\displaystyle \textbf{Question 34. }\text{Find the roots of the quadratic equation}$
$\displaystyle \sqrt{2}x^{2}+7x+5\sqrt{2}=0. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{2}x^{2}+7x+5\sqrt{2}=0$
$\displaystyle \text{Here, }a=\sqrt{2},\ b=7,\ c=5\sqrt{2}$
$\displaystyle D=b^{2}-4ac$
$\displaystyle =(7)^{2}-4\times\sqrt{2}\times5\sqrt{2}$
$\displaystyle \Rightarrow D=49-40=9$
$\displaystyle \text{Since, }D>0$
$\displaystyle \therefore \text{Roots are real and unequal.}$
$\displaystyle \text{By quadratic formula,}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}=\frac{-7\pm3}{2\sqrt{2}}$
$\displaystyle \Rightarrow x=\frac{-7+3}{2\sqrt{2}}\text{ or }\frac{-7-3}{2\sqrt{2}}$
$\displaystyle \Rightarrow x=\frac{-4}{2\sqrt{2}}\text{ or }\frac{-10}{2\sqrt{2}}$
$\displaystyle \Rightarrow x=-\sqrt{2}\text{ or }-\frac{5}{\sqrt{2}}$
$\displaystyle \therefore \text{Roots are }-\sqrt{2}\text{ and }-\frac{5}{\sqrt{2}}.$
$\\$

$\displaystyle \textbf{Question 35. }\text{Solve the following quadratic equation for }x:$
$\displaystyle 4x^{2}-4a^{2}x+(a^{4}-b^{4})=0 \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle 4x^{2}-4a^{2}x+(a^{4}-b^{4})=0$
$\displaystyle \Rightarrow x=\frac{4a^{2}\pm\sqrt{16a^{4}-4\times4\times(a^{4}-b^{4})}}{2\times4}$
$\displaystyle \left[\because x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}\right]$
$\displaystyle \Rightarrow x=\frac{4a^{2}\pm\sqrt{16b^{4}}}{2\times4}$
$\displaystyle \Rightarrow x=\frac{4a^{2}\pm4b^{2}}{2\times4}=\frac{a^{2}\pm b^{2}}{2}$
$\displaystyle \Rightarrow x=\frac{a^{2}+b^{2}}{2},\ \frac{a^{2}-b^{2}}{2}$
$\\$

$\displaystyle \textbf{Question 36. }\text{Solve for }x:$
$\displaystyle \frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9},\ x\neq \frac{3}{2},5 \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is}$
$\displaystyle \frac{1}{2x-3}+\frac{1}{x-5}=1\frac{1}{9}$
$\displaystyle \Rightarrow \frac{1}{2x-3}+\frac{1}{x-5}=\frac{10}{9}$
$\displaystyle \Rightarrow \frac{(x-5)+(2x-3)}{(2x-3)(x-5)}=\frac{10}{9}$
$\displaystyle \Rightarrow \frac{x-5+2x-3}{2x^{2}-13x+15}=\frac{10}{9}$
$\displaystyle \Rightarrow 9(3x-8)=10(2x^{2}-13x+15)$
$\displaystyle \Rightarrow 27x-72=20x^{2}-130x+150$
$\displaystyle \Rightarrow 20x^{2}-157x+222=0$
$\displaystyle \text{Here, }a=20,\ b=-157,\ c=222$
$\displaystyle \text{Now, }D=b^{2}-4ac$
$\displaystyle D=(-157)^{2}-4\times20\times222=24649-17760=6889$
$\displaystyle \sqrt{D}=\sqrt{6889}=83$
$\displaystyle \text{By quadratic formula,}$
$\displaystyle x=\frac{-b\pm\sqrt{D}}{2a}$
$\displaystyle x=\frac{157+83}{40},\ \frac{157-83}{40}$
$\displaystyle x=\frac{240}{40},\ \frac{74}{40}$
$\displaystyle \therefore x=6,\ \frac{37}{20}$
$\\$

$\displaystyle \textbf{Question 37. }\text{Find the value of }k\text{ for which the equation}$
$\displaystyle x^{2}+k(2x+k-1)=0\text{ has real and equal roots.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle x^{2}+k(2x+k-1)=0$
$\displaystyle \Rightarrow x^{2}+2kx+k^{2}-k=0$
$\displaystyle \text{Here, }a=1,\ b=2k,\ c=k^{2}-k$
$\displaystyle \text{Since, the roots are real and equal}$
$\displaystyle D=b^{2}-4ac=0$
$\displaystyle \Rightarrow (2k)^{2}-4\times1\times(k^{2}-k)=0$
$\displaystyle \Rightarrow 4k^{2}-4k^{2}+4k=0$
$\displaystyle \Rightarrow 4k=0$
$\displaystyle \Rightarrow k=0$
$\\$

$\displaystyle \textbf{Question 38. }\text{Find the value of }p,\text{ for which one root of the}$
$\displaystyle \text{quadratic equation }px^{2}-14x+8=0\text{ is }6\text{ times the other.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given quadratic equation is}$
$\displaystyle px^{2}-14x+8=0$
$\displaystyle \text{Here }a=p,\ b=-14,\ c=8$
$\displaystyle \text{Let }\alpha,\beta\text{ be the two roots of this equation such that }\alpha=6\beta$
$\displaystyle \text{Now, sum of roots}=-\frac{b}{a}$
$\displaystyle \alpha+\beta=-\frac{b}{a}$
$\displaystyle 6\beta+\beta=\frac{14}{p}$
$\displaystyle 7\beta=\frac{14}{p}\Rightarrow \beta=\frac{2}{p} \qquad (i)$
$\displaystyle \text{Also, product of roots}=\frac{c}{a}$
$\displaystyle \alpha\beta=\frac{c}{a}$
$\displaystyle 6\beta\cdot\beta=\frac{8}{p}$
$\displaystyle \beta^{2}=\frac{4}{3p} \qquad (ii)$
$\displaystyle \text{Solving equations }(i)\text{ and }(ii)\text{ to eliminate }\beta,\text{ we get}$
$\displaystyle \left(\frac{2}{p}\right)^{2}=\frac{4}{3p}$
$\displaystyle \Rightarrow \frac{4}{p^{2}}=\frac{4}{3p}$
$\displaystyle \Rightarrow p^{2}-3p=0$
$\displaystyle \Rightarrow p(p-3)=0\Rightarrow p=0,\ 3$
$\displaystyle \text{If }p=0,\text{ then the given equation will not remain quadratic.}$
$\displaystyle \text{So, }p=0\text{ is rejected.}$
$\displaystyle \therefore \text{The value of }p\text{ is }3.$
$\\$

$\displaystyle \textbf{Question 39. }\text{Find the value of }k\text{ for which the roots of the}$
$\displaystyle \text{quadratic equation }2x^{2}+kx+8=0\text{ will have equal value.} \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation}$
$\displaystyle 2x^{2}+kx+8=0$
$\displaystyle \text{Roots of quadratic equation are equal.}$
$\displaystyle \Rightarrow D=0$
$\displaystyle \Rightarrow b^{2}-4ac=0$
$\displaystyle \Rightarrow k^{2}-4(2)(8)=0$
$\displaystyle \Rightarrow k^{2}-64=0$
$\displaystyle \Rightarrow k^{2}=64$
$\displaystyle \Rightarrow k=\pm 8$
$\\$

$\displaystyle \textbf{Question 40. }\text{If the equation }(1+m^{2})x^{2}+2mcx+c^{2}-a^{2}=0$
$\displaystyle \text{has equal roots, then show that }c^{2}=a^{2}(1+m^{2}). \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is}$
$\displaystyle (1+m^{2})x^{2}+2mcx+c^{2}-a^{2}=0$
$\displaystyle \text{Here }A=(1+m^{2}),\ B=2mc,\ C=c^{2}-a^{2}$
$\displaystyle \text{This equation has equal roots, }\therefore D=0$
$\displaystyle \Rightarrow B^{2}-4AC=0$
$\displaystyle \Rightarrow (2mc)^{2}-4(1+m^{2})(c^{2}-a^{2})=0$
$\displaystyle \Rightarrow 4m^{2}c^{2}-4(c^{2}-a^{2}+m^{2}c^{2}-m^{2}a^{2})=0$
$\displaystyle \Rightarrow 4m^{2}c^{2}-4c^{2}+4a^{2}-4m^{2}c^{2}+4m^{2}a^{2}=0$
$\displaystyle \Rightarrow 4c^{2}-4a^{2}-4m^{2}a^{2}=0$
$\displaystyle \Rightarrow c^{2}-a^{2}-m^{2}a^{2}=0$
$\displaystyle \Rightarrow c^{2}=a^{2}+m^{2}a^{2}$
$\displaystyle \Rightarrow c^{2}=a^{2}(1+m^{2})$
$\\$

$\displaystyle \textbf{Question 41. }\text{If the roots of the equation }(a^{2}+b^{2})x^{2}-2(ac+bd)x$
$\displaystyle +(c^{2}+d^{2})=0\text{ are equal, prove that }\frac{a}{b}=\frac{c}{d}. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle (a^{2}+b^{2})x^{2}-2(ac+bd)x+(c^{2}+d^{2})=0$
$\displaystyle \text{Here, }A=(a^{2}+b^{2}),\ B=-2(ac+bd),\ C=c^{2}+d^{2}$
$\displaystyle \text{Roots are equal. }\therefore D=0$
$\displaystyle \Rightarrow B^{2}-4AC=0$
$\displaystyle \Rightarrow B^{2}=4AC$
$\displaystyle \Rightarrow [-2(ac+bd)]^{2}=4(a^{2}+b^{2})(c^{2}+d^{2})$
$\displaystyle \Rightarrow 4(a^{2}c^{2}+b^{2}d^{2}+2acbd)=4(a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2})$
$\displaystyle \Rightarrow a^{2}c^{2}+b^{2}d^{2}+2acbd=a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2}$
$\displaystyle \Rightarrow 2acbd=a^{2}d^{2}+b^{2}c^{2}$
$\displaystyle \Rightarrow a^{2}d^{2}+b^{2}c^{2}-2abcd=0$
$\displaystyle \Rightarrow (ad-bc)^{2}=0$
$\displaystyle \Rightarrow ad=bc$
$\displaystyle \Rightarrow \frac{a}{b}=\frac{c}{d}$
$\displaystyle \text{Hence proved.}$
$\\$

$\displaystyle \textbf{Question 42. }\text{If }ad\neq bc,\text{ then prove that the equation}$
$\displaystyle (a^{2}+b^{2})x^{2}+2(ac+bd)x+(c^{2}+d^{2})=0\text{ has no real roots.} \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is}$
$\displaystyle (a^{2}+b^{2})x^{2}+2(ac+bd)x+(c^{2}+d^{2})=0$
$\displaystyle \text{Here }A=(a^{2}+b^{2}),\ B=2(ac+bd),\ C=(c^{2}+d^{2})$
$\displaystyle \text{To find the nature of its roots, we find the discriminant.}$
$\displaystyle D=B^{2}-4AC$
$\displaystyle =[2(ac+bd)]^{2}-4(a^{2}+b^{2})(c^{2}+d^{2})$
$\displaystyle =4(a^{2}c^{2}+b^{2}d^{2}+2abcd)-4(a^{2}c^{2}+a^{2}d^{2}+b^{2}c^{2}+b^{2}d^{2})$
$\displaystyle =4a^{2}c^{2}+4b^{2}d^{2}+8abcd-4a^{2}c^{2}-4a^{2}d^{2}-4b^{2}c^{2}-4b^{2}d^{2}$
$\displaystyle =-4(a^{2}d^{2}+b^{2}c^{2}-2abcd)$
$\displaystyle D=-4(ad-bc)^{2}$
$\displaystyle \text{It is given that }ad\neq bc$
$\displaystyle \Rightarrow ad-bc\neq0$
$\displaystyle \Rightarrow (ad-bc)^{2}>0\text{ for all values of }a,b,c\text{ and }d$
$\displaystyle \Rightarrow -4(ad-bc)^{2}<0$
$\displaystyle \Rightarrow D<0$
$\displaystyle \text{Hence the given equation has no real roots.}$
$\\$

$\displaystyle \textbf{Question 43. }\text{If the roots of the equation }(c^{2}-ab)x^{2}-2(a^{2}-bc)x$
$\displaystyle +b^{2}-ac=0\text{ in }x\text{ are equal, then show that either }a=0\text{ or }a^{3}+b^{3}+c^{3}=3abc. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given quadratic equation is}$
$\displaystyle (c^{2}-ab)x^{2}-2(a^{2}-bc)x+b^{2}-ac=0$
$\displaystyle \text{Here, }A=(c^{2}-ab),\ B=-2(a^{2}-bc)\text{ and }C=(b^{2}-ac)$
$\displaystyle \text{The roots are equal}$
$\displaystyle \Rightarrow \text{Discriminant}=0$
$\displaystyle \Rightarrow B^{2}-4AC=0$
$\displaystyle \Rightarrow \{-2(a^{2}-bc)\}^{2}-4(c^{2}-ab)(b^{2}-ac)=0$
$\displaystyle \Rightarrow 4(a^{2}-bc)^{2}-4(c^{2}-ab)(b^{2}-ac)=0$
$\displaystyle \Rightarrow a^{4}-2a^{2}bc+b^{2}c^{2}-c^{2}b^{2}+ac^{3}+ab^{3}-a^{2}bc=0$
$\displaystyle \Rightarrow a[a^{3}-2abc+c^{3}+b^{3}-abc]=0$
$\displaystyle \Rightarrow a[a^{3}+b^{3}+c^{3}-3abc]=0$
$\displaystyle \Rightarrow \text{either }a=0\text{ or }a^{3}+b^{3}+c^{3}=3abc$
$\displaystyle \text{Hence, if roots of the given quadratic equation are equal, then either }a=0\text{ or }a^{3}+b^{3}+c^{3}=3abc.$
$\\$

$\displaystyle \textbf{Question 44. }\text{If the roots of the quadratic equation }(a-b)x^{2}$
$\displaystyle +(b-c)x+(c-a)=0\text{ are equal, prove that }2a=b+c. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Here, }A=(a-b),\ B=(b-c),\ C=(c-a)$
$\displaystyle \text{For equal roots}$
$\displaystyle D=0$
$\displaystyle \Rightarrow B^{2}-4AC=0$
$\displaystyle \Rightarrow (b-c)^{2}-4(a-b)(c-a)=0$
$\displaystyle \Rightarrow b^{2}+c^{2}-2bc-4ac+4a^{2}+4bc-4ab=0$
$\displaystyle \Rightarrow (b+c-2a)^{2}=0$
$\displaystyle \Rightarrow 2a=b+c$
$\\$

$\displaystyle \textbf{Question 45. }\text{If the roots of the quadratic equation}$
$\displaystyle (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\displaystyle \text{are equal, then show that }a=b=c. \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle (x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)=0$
$\displaystyle \Rightarrow x^{2}-bx-ax+ab+x^{2}-cx-bx+bc+x^{2}-ax-cx+ac=0$
$\displaystyle \Rightarrow 3x^{2}-2ax-2bx-2cx+ab+bc+ca=0$
$\displaystyle \Rightarrow 3x^{2}-2(a+b+c)x+ab+bc+ca=0$
$\displaystyle \text{Here, }A=3,\ B=-2(a+b+c),\ C=ab+bc+ca$
$\displaystyle \text{For equal roots}$
$\displaystyle D=B^{2}-4AC=0$
$\displaystyle \Rightarrow 4(a+b+c)^{2}-12(ab+bc+ca)=0$
$\displaystyle \Rightarrow 4(a^{2}+b^{2}+c^{2}-ab-bc-ca)=0$
$\displaystyle \Rightarrow 2[a^{2}+a^{2}+b^{2}+b^{2}+c^{2}+c^{2}-2ab-2bc-2ca]=0$
$\displaystyle \Rightarrow 2[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]=0$
$\displaystyle \Rightarrow a-b=0,\ b-c=0,\ c-a=0$
$\displaystyle \Rightarrow a=b,\ b=c,\ c=a$
$\displaystyle \therefore a=b=c$
$\\$

$\displaystyle \textbf{Question 46. }\text{Solve for }x:$
$\displaystyle \sqrt{3}x^{2}-2x-8\sqrt{3}=0 \hspace{0.2cm}\text{[CBSE 2017]}$
$\displaystyle \text{Answer:}$
$\displaystyle \ \sqrt{3}x^{2}-2x-8\sqrt{3}=0$
$\displaystyle \sqrt{3}x^{2}-6x+4x-8\sqrt{3}=0$
$\displaystyle \sqrt{3}x(x-2\sqrt{3})+4(x-2\sqrt{3})=0$
$\displaystyle (\sqrt{3}x+4)(x-2\sqrt{3})=0$
$\displaystyle \therefore x=-\frac{4}{\sqrt{3}},\ 2\sqrt{3}$
$\\$

$\displaystyle \textbf{Question 47. }\text{Solve for }x:$
$\displaystyle \frac{x+3}{x+2}=\frac{3x-7}{2x-3},\ x\neq -2,\ \frac{3}{2} \hspace{0.2cm}\text{[CBSE 2017(C)]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+3}{x+2}=\frac{3x-7}{2x-3}$
$\displaystyle \text{Cross multiplying, we get}$
$\displaystyle (x+3)(2x-3)=(3x-7)(x+2)$
$\displaystyle \Rightarrow 2x^{2}-3x+6x-9=3x^{2}+6x-7x-14$
$\displaystyle \Rightarrow x^{2}-4x-5=0$
$\displaystyle \Rightarrow (x-5)(x+1)=0$
$\displaystyle \therefore x=5\ \text{or}\ x=-1$
$\\$

$\displaystyle \textbf{Question 48. }\text{Solve for }x: \sqrt{2x+9}+x=13. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \sqrt{2x+9}+x=13$
$\displaystyle \Rightarrow \sqrt{2x+9}=13-x$
$\displaystyle \text{Squaring both sides, we get}$
$\displaystyle 2x+9=(13-x)^{2}$
$\displaystyle \Rightarrow 2x+9=169-26x+x^{2}$
$\displaystyle \Rightarrow x^{2}-28x+160=0$
$\displaystyle \Rightarrow (x-20)(x-8)=0$
$\displaystyle \therefore x=20\ \text{or}\ x=8$
$\displaystyle [\text{as }x=20\text{ does not satisfy the equation}]$
$\displaystyle \therefore x=8$
$\\$

$\displaystyle \textbf{Question 49. }\text{A two digit number is four times the sum of the digits. It is also equal to }$
$\displaystyle 3\text{ times the product of} \text{ digits. Find the number.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let ones digit of number}=x$
$\displaystyle \text{Let tens digit of number}=y$
$\displaystyle \therefore \text{Number will be }10y+x$
$\displaystyle \text{According to question,}$
$\displaystyle 10y+x=4(x+y)$
$\displaystyle \Rightarrow 10y+x=4x+4y$
$\displaystyle \Rightarrow 3x-6y=0$
$\displaystyle \Rightarrow x-2y=0 \qquad (i)$
$\displaystyle \text{and}$
$\displaystyle 10y+x=3y^{2} \qquad (ii)$
$\displaystyle \text{Putting }x=2y\text{ from }(i)\text{ in }(ii),\ \text{we get}$
$\displaystyle 10y+2y=3y^{2}$
$\displaystyle \Rightarrow 12y=6y^{2}$
$\displaystyle \Rightarrow y(y-2)=0$
$\displaystyle \therefore y=0\ \text{or}\ y=2$
$\displaystyle [y=0\text{ rejected}]$
$\displaystyle \therefore y=2$
$\displaystyle \text{and }x=2y=4$
$\displaystyle \therefore \text{Required number}=10(2)+4=24$
$\\$

$\displaystyle \textbf{Question 50. }\text{Solve for }x:$
$\displaystyle \frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2};\ x\neq 1,-2,2 \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+1}{x-1}+\frac{x-2}{x+2}=4-\frac{2x+3}{x-2}$
$\displaystyle \Rightarrow \frac{x+1}{x-1}+\frac{x-2}{x+2}+\frac{2x+3}{x-2}=4$
$\displaystyle \Rightarrow \frac{(x+1)(x+2)(x-2)+(x-2)^{2}(x-1)+(2x+3)(x-1)(x+2)}{(x-1)(x+2)(x-2)}=4$
$\displaystyle \Rightarrow (x+1)(x^{2}-4)+(x-1)(x^{2}+4-4x)+(2x+3)(x^{2}+x-2)=4(x^{3}-4x-x^{2}+4)$
$\displaystyle \Rightarrow x^{3}-4x+x^{2}-4+x^{3}+4x-4x^{2}-x^{2}-4+4x+2x^{3}+2x^{2}-4x+3x^{2}+3x-6=4(x^{3}-4x-x^{2}+4)$
$\displaystyle \Rightarrow 4x^{3}+x^{2}+3x-14=4x^{3}-4x^{2}-16x+16$
$\displaystyle \Rightarrow 5x^{2}+19x-30=0$
$\displaystyle \Rightarrow 5x^{2}+25x-6x-30=0$
$\displaystyle \Rightarrow 5x(x+5)-6(x+5)=0$
$\displaystyle \Rightarrow (x+5)(5x-6)=0$
$\displaystyle \Rightarrow x+5=0\text{ or }5x-6=0$
$\displaystyle \Rightarrow x=-5\text{ or }x=\frac{6}{5}$
$\displaystyle \text{Thus, solutions of given equation are }x=-5\text{ and }x=\frac{6}{5}.$
$\\$

$\displaystyle \textbf{Question 51. }\text{Solve the following quadratic equation for }x:$
$\displaystyle x^{2}+\left(\frac{a}{a+b}+\frac{a+b}{a}\right)x+1=0 \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle x^{2}+\left(\frac{a}{a+b}+\frac{a+b}{a}\right)x+1=0$
$\displaystyle \Rightarrow x^{2}+\left(\frac{a}{a+b}+\frac{a+b}{a}\right)x+\left(\frac{a}{a+b}\right)\left(\frac{a+b}{a}\right)=0$
$\displaystyle \Rightarrow x^{2}+\left(\frac{a}{a+b}\right)x+\left(\frac{a+b}{a}\right)x+\left(\frac{a}{a+b}\right)\left(\frac{a+b}{a}\right)=0$
$\displaystyle \Rightarrow x\left[x+\frac{a}{a+b}\right]+\frac{a+b}{a}\left[x+\frac{a}{a+b}\right]=0$
$\displaystyle \Rightarrow \left(x+\frac{a}{a+b}\right)\left(x+\frac{a+b}{a}\right)=0$
$\displaystyle \Rightarrow x+\frac{a}{a+b}=0\text{ or }x+\frac{a+b}{a}=0$
$\displaystyle \Rightarrow x=-\frac{a}{a+b}\text{ or }x=-\frac{a+b}{a}$
$\\$

$\displaystyle \textbf{Question 52. }\text{Solve for }x:$
$\displaystyle \frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3},\qquad x\neq 1,2,3 \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}=\frac{2}{3},\ x\neq1,2,3$
$\displaystyle \Rightarrow \frac{x-3+x-1}{(x-1)(x-2)(x-3)}=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{2(x-2)}{(x-1)(x-2)(x-3)}=\frac{2}{3}$
$\displaystyle \Rightarrow \frac{1}{(x-1)(x-3)}=\frac{1}{3}$
$\displaystyle \Rightarrow 3=(x-1)(x-3)$
$\displaystyle \Rightarrow x^{2}-4x+3=3$
$\displaystyle \Rightarrow x^{2}-4x=0$
$\displaystyle \Rightarrow x(x-4)=0$
$\displaystyle \therefore x=0\text{ or }x=4$
$\\$

$\displaystyle \textbf{Question 53. }\text{Three consecutive natural numbers are such that the square of the middle number}$
$\displaystyle \text{exceeds the difference of the squares of the other two by }60. \text{ Find the numbers.}$
$\displaystyle \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let three consecutive natural numbers be }x-1,\ x\text{ and }x+1.$
$\displaystyle \text{According to question,}$
$\displaystyle x^{2}-[(x+1)^{2}-(x-1)^{2}]=60$
$\displaystyle \Rightarrow x^{2}-[(x+1-x+1)(x+1+x-1)]=60$
$\displaystyle \Rightarrow x^{2}-4x-60=0$
$\displaystyle \Rightarrow x^{2}-10x+6x-60=0$
$\displaystyle \Rightarrow x(x-10)+6(x-10)=0$
$\displaystyle \Rightarrow (x+6)(x-10)=0$
$\displaystyle \Rightarrow x=10$
$\displaystyle (x=-6,\ \text{rejected})$
$\displaystyle \therefore \text{The numbers are }9,\ 10\text{ and }11.$
$\\$

$\displaystyle \textbf{Question 54. }\text{The time taken by a person to cover }150\text{ km was }2\frac{1}{2}\text{ hours more than the}$
$\displaystyle \text{time taken in the return journey. If he returned at a speed of }10\text{ km/hour more than}$
$\displaystyle \text{the speed while going, find the speed per hour in each direction.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }t_{1}\text{ and }t_{2}\text{ be the time taken in going and returning respectively.}$
$\displaystyle \text{Also, }v\text{ and }v+10\text{ be the speed in going and returning respectively.}$
$\displaystyle \text{As, Velocity}=\frac{\text{Distance}}{\text{Time}}$
$\displaystyle \Rightarrow \text{Time}=\frac{\text{Distance}}{\text{Velocity}}$
$\displaystyle \therefore t_{1}=\frac{150}{v}\text{ and }t_{2}=\frac{150}{v+10}$
$\displaystyle \text{According to question, }t_{1}-t_{2}=2\frac{1}{2}$
$\displaystyle \frac{150}{v}-\frac{150}{v+10}=\frac{5}{2}$
$\displaystyle \Rightarrow 150\left[\frac{1}{v}-\frac{1}{v+10}\right]=\frac{5}{2}$
$\displaystyle \Rightarrow \frac{1}{v}-\frac{1}{v+10}=\frac{1}{60}$
$\displaystyle \Rightarrow 60[v+10-v]=v(v+10)$
$\displaystyle \Rightarrow v^{2}+10v-600=0$
$\displaystyle \Rightarrow (v-20)(v+30)=0$
$\displaystyle \therefore v=20\text{ or }v=-30\text{ (rejected)}$
$\displaystyle \text{Hence, velocity in going is }20\text{ km/hour and in returning }30\text{ km/hour.}$
$\\$

$\displaystyle \textbf{Question 55. }\text{A rectangular park is to be designed so that its breadth is }3\text{ m less than its}$
$\displaystyle \text{length. Its area is to be }4\text{ square metres more than the area of a park that has already}$
$\displaystyle \text{been made in the shape of an isosceles triangle with its base as the breadth of the}$
$\displaystyle \text{rectangular park and altitude }12\text{ m. Find the length and breadth of the rectangular}$
$\displaystyle \text{park.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let length of rectangular park be }x\text{ m and breadth be }(x-3)\text{ m.}$
$\displaystyle \text{Base of isosceles }\triangle=(x-3)\text{ m}$
$\displaystyle \text{Altitude of }\triangle=12\text{ m}$
$\displaystyle \text{According to question,}$
$\displaystyle \text{Area of rectangular park}=\text{Area of }\triangle+4$
$\displaystyle \Rightarrow x(x-3)=\frac{1}{2}(x-3)\times12+4$
$\displaystyle \Rightarrow x(x-3)=6(x-3)+4$
$\displaystyle \Rightarrow x^{2}-3x=6x-18+4$
$\displaystyle \Rightarrow x^{2}-9x+14=0$
$\displaystyle \Rightarrow (x-7)(x-2)=0$
$\displaystyle \Rightarrow x=7\text{ or }x=2$
$\displaystyle \text{But }x=2\text{ is rejected otherwise breadth will be negative, which is not possible.}$
$\displaystyle \therefore \text{Length of rectangular park is }7\text{ m and breadth is }4\text{ m.}$
$\\$

$\displaystyle \textbf{Question 56. }\text{A pole is to be erected at a point on the boundary of a circular park of diameter}$
$\displaystyle 17\text{ m in such a way that the differences of its distances from the diametrically}$
$\displaystyle \text{opposite fixed gates A and B on the boundary is }7\text{ metres. Find the distances from}$
$\displaystyle \text{the two gates where the pole is to be erected.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }P\text{ be the position of the pole.}$
$\displaystyle \angle APB=90^\circ\text{ (angle in a semicircle)}$
$\displaystyle \text{By Pythagoras theorem,}$
$\displaystyle AB^{2}=AP^{2}+PB^{2}$
$\displaystyle 17^{2}=AP^{2}+PB^{2} \qquad (i)$
$\displaystyle \text{Also, }AP-PB=7 \qquad (ii)$
$\displaystyle \text{Squaring both sides, we get}$
$\displaystyle (AP-PB)^{2}=49$
$\displaystyle AP^{2}+PB^{2}-2AP\cdot PB=49 \qquad (iii)$
$\displaystyle \text{From }(i)\text{ and }(iii),\ \text{we have}$
$\displaystyle 17^{2}-2AP\cdot PB=49$
$\displaystyle \Rightarrow 2AP\cdot PB=289-49$
$\displaystyle \Rightarrow AP\cdot PB=\frac{240}{2}$
$\displaystyle \Rightarrow AP\cdot PB=120 \qquad (iv)$
$\displaystyle \text{From }(ii)\text{ and }(iv),\ \text{we have}$
$\displaystyle 120=PB(7+PB)$
$\displaystyle \text{Let }PB=x$
$\displaystyle 120=x(7+x)$
$\displaystyle \Rightarrow x^{2}+7x-120=0$
$\displaystyle \Rightarrow (x-8)(x+15)=0$
$\displaystyle \Rightarrow x=8\text{ or }x=-15\text{ (rejected)}$
$\displaystyle \therefore PB=x=8\text{ m}$
$\displaystyle AP=7+PB=7+8=15\text{ m}$
$\\$

$\displaystyle \textbf{Question 57. }\text{The denominator of a fraction is one more than twice its numerator. }$
$\displaystyle \text{If the sum of the fraction and its reciprocal is }2\frac{16}{21},\text{ find the fraction.} \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the numerator}=x$
$\displaystyle \text{Then, denominator}=2x+1$
$\displaystyle \text{Required fraction}=\frac{x}{2x+1}$
$\displaystyle \text{According to question,}$
$\displaystyle \frac{x}{2x+1}+\frac{2x+1}{x}=2\frac{16}{21}$
$\displaystyle \text{Consider }\frac{x}{2x+1}=y,\text{ then above equation becomes,}$
$\displaystyle y+\frac{1}{y}=2\frac{16}{21}$
$\displaystyle \Rightarrow \frac{y^{2}+1}{y}=\frac{58}{21}$
$\displaystyle \Rightarrow 21y^{2}-58y+21=0$
$\displaystyle \Rightarrow 21y^{2}-49y-9y+21=0$
$\displaystyle \Rightarrow 7y(3y-7)-3(3y-7)=0$
$\displaystyle \Rightarrow (7y-3)(3y-7)=0$
$\displaystyle \Rightarrow 7y-3=0\text{ or }3y-7=0$
$\displaystyle \Rightarrow y=\frac{3}{7}\text{ or }y=\frac{7}{3}$
$\displaystyle \Rightarrow \frac{x}{2x+1}=\frac{3}{7}\text{ or }\frac{x}{2x+1}=\frac{7}{3}$
$\displaystyle \Rightarrow x=3\text{ or }x=-\frac{7}{11}\text{ (rejected)}$
$\displaystyle \text{So, fraction}=\frac{3}{2\times3+1}=\frac{3}{7}$
$\\$

$\displaystyle \textbf{Question 58. }\text{Solve the given quadratic equation for }x:$
$\displaystyle 9x^{2}-9(a+b)x+(2a^{2}+5ab+2b^{2})=0. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^{2}-9(a+b)x+(2a^{2}+5ab+2b^{2})=0$
$\displaystyle \text{Comparing with }Ax^{2}+Bx+C=0,$
$\displaystyle A=9,\ B=-9(a+b),\ C=2a^{2}+5ab+2b^{2}$
$\displaystyle \text{Roots of above quadratic equation are given by quadratic formula,}$
$\displaystyle x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}$
$\displaystyle x=\frac{-[-9(a+b)]\pm\sqrt{[-9(a+b)]^{2}-4(9)(2a^{2}+5ab+2b^{2})}}{2(9)}$
$\displaystyle x=\frac{9(a+b)\pm3\sqrt{9a^{2}+9b^{2}+18ab-8a^{2}-8b^{2}-20ab}}{18}$
$\displaystyle x=\frac{9(a+b)\pm3\sqrt{a^{2}+b^{2}-2ab}}{18}$
$\displaystyle x=\frac{9(a+b)\pm3(a-b)}{18}$
$\displaystyle \therefore x=\frac{9a+9b+3a-3b}{18}\text{ and }x=\frac{9a+9b-3a+3b}{18}$
$\displaystyle \therefore x=\frac{2a+b}{3}\text{ and }x=\frac{a+2b}{3}$
$\\$

$\displaystyle \textbf{Question 59. }\text{Solve for }x\text{ (in terms of }a\text{ and }b):$
$\displaystyle \frac{a}{x-b}+\frac{b}{x-a}=2,\ x\neq a,b \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{a}{x-b}+\frac{b}{x-a}=2;\ x\neq a,b$
$\displaystyle \Rightarrow \frac{a(x-a)+b(x-b)}{(x-a)(x-b)}=2$
$\displaystyle \Rightarrow ax-a^{2}+bx-b^{2}=2(x^{2}-ax-bx+ab)$
$\displaystyle \Rightarrow ax+bx-a^{2}-b^{2}=2x^{2}-2ax-2bx+2ab$
$\displaystyle \Rightarrow 2x^{2}-3ax-3bx+a^{2}+b^{2}+2ab=0$
$\displaystyle \Rightarrow 2x^{2}-3(a+b)x+(a+b)^{2}=0$
$\displaystyle \text{Comparing with }Ax^{2}+Bx+C=0,\ \text{we get}$
$\displaystyle A=2,\ B=-3(a+b),\ C=(a+b)^{2}$
$\displaystyle \text{By quadratic formula,}$
$\displaystyle x=\frac{-B\pm\sqrt{B^{2}-4AC}}{2A}$
$\displaystyle \Rightarrow x=\frac{3(a+b)\pm\sqrt{9(a+b)^{2}-8(a+b)^{2}}}{4}$
$\displaystyle \Rightarrow x=\frac{3(a+b)\pm(a+b)}{4}$
$\displaystyle \Rightarrow x=\frac{3(a+b)+(a+b)}{4}\text{ or }x=\frac{3(a+b)-(a+b)}{4}$
$\displaystyle \Rightarrow x=a+b\text{ or }x=\frac{a+b}{2}$
$\\$

$\displaystyle \textbf{Question 60. }\text{If }-5\text{ is a root of the quadratic equation}$
$\displaystyle 2x^{2}+px-15=0\text{ and the quadratic equation}$
$\displaystyle p(x^{2}+x)+k=0\text{ has equal roots, find the value of }k. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }-5\text{ is the root of the quadratic equation }2x^{2}+px-15=0$
$\displaystyle \Rightarrow 2(-5)^{2}+p(-5)-15=0$
$\displaystyle \Rightarrow 50-5p-15=0\Rightarrow 35-5p=0$
$\displaystyle \Rightarrow 5p=35\Rightarrow p=7$
$\displaystyle \text{Now, given that equation }p(x^{2}+x)+k=0\text{ has equal roots.}$
$\displaystyle \text{i.e. }7x^{2}+7x+k=0\text{ has equal roots.}$
$\displaystyle \Rightarrow 7^{2}-4\times7\times k=0$
$\displaystyle \left[\because \text{For equal roots, }D=0,\ \text{i.e. }b^{2}-4ac=0\right]$
$\displaystyle \Rightarrow 7(7-4k)=0$
$\displaystyle \Rightarrow k=\frac{7}{4}$
$\\$

$\displaystyle \textbf{Question 61. }\text{If }x=\frac{2}{3}\text{ and }x=-3\text{ are roots of the quadratic equation } \\ ax^{2}+7x+b=0,\text{ find the values of }a\text{ and }b. \hspace{0.2cm}\text{[CBSE 2016]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }ax^{2}+7x+b=0 \qquad (i)$
$\displaystyle a\left(\frac{2}{3}\right)^{2}+7\left(\frac{2}{3}\right)+b=0$
$\displaystyle \left[\because x=\frac{2}{3}\text{ is the root of equation }(i)\right]$
$\displaystyle \Rightarrow \frac{4a}{9}+\frac{14}{3}+b=0$
$\displaystyle \Rightarrow \frac{4a+42+9b}{9}=0$
$\displaystyle \Rightarrow 4a+9b+42=0 \qquad (ii)$
$\displaystyle \text{Also }a(-3)^{2}+7(-3)+b=0$
$\displaystyle \left[\because x=-3\text{ is the root of equation }(i)\right]$
$\displaystyle \Rightarrow 9a+b-21=0$
$\displaystyle \Rightarrow 9a+b=21 \qquad (iii)$
$\displaystyle \text{Putting value of }b\text{ from }(iii)\text{ in }(ii),\ \text{we get}$
$\displaystyle 4a+9(21-9a)+42=0$
$\displaystyle \Rightarrow 4a+189-81a+42=0$
$\displaystyle \Rightarrow 77a=231$
$\displaystyle \Rightarrow a=3$
$\displaystyle \text{Putting }a=3\text{ in }(iii),$
$\displaystyle 27+b=21$
$\displaystyle \Rightarrow b=-6$
$\\$