Class 10 – Quadratic Equations – ICSE Board Problems (1)

$\displaystyle \text{Question 1: Car A travels }x\text{ km for every liter of petrol, while car B travels } \\ (x+5)\text{ km for every liter of petrol.}$
$\displaystyle \text{(i) Write down the number of liters of petrol used by car A and car B in covering a distance of } \\ 400\text{ km.}$
$\displaystyle \text{(ii) If car A uses }4\text{ liters of petrol more than car B in covering the }400\text{ km,} \\ \text{write down an equation in }x\text{ and solve it to determine the number of liters of petrol used} \\ \text{by car B for the journey.}\hfill \text{[ICSE 1997]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) No. of liters of petrol used by car A }=\frac{400}{x}\text{ liters}$
$\displaystyle \text{No. of liters of petrol used by car B }=\frac{400}{x+5}\text{ liters}$
$\displaystyle \text{(ii) Given, }\frac{400}{x}-\frac{400}{x+5}=4$
$\displaystyle \frac{400(x+5)-400x}{x(x+5)}=4$
$\displaystyle \frac{2000}{x(x+5)}=4$
$\displaystyle 2000=4x(x+5)$
$\displaystyle 500=x^2+5x$
$\displaystyle x^2+5x-500=0$
$\displaystyle (x+25)(x-20)=0$
$\displaystyle x=-25\text{ or }x=20$
$\displaystyle \text{Since distance per litre cannot be negative, }x=20$
$\displaystyle \therefore \text{No. of liters of petrol used by car B }=\frac{400}{20+5}=16\text{ liters}$
$\\$

$\displaystyle \text{Question 2: By increasing the speed of a car by }10\text{ km/hr, the time of journey for a} \\ \text{distance of } 72\text{ km is reduced by }36\text{ minutes. Find the original speed of the car.}\hfill \text{[ICSE 2005]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the original speed of the car }=x\text{ km/hr}$
$\displaystyle \text{Time taken to cover }72\text{ km }=\frac{72}{x}\text{ hrs}$
$\displaystyle \text{New speed of the car }=(x+10)\text{ km/hr}$
$\displaystyle \text{New time taken to cover }72\text{ km }=\frac{72}{x+10}\text{ hrs}$
$\displaystyle 36\text{ minutes}=\frac{36}{60}\text{ hrs}=\frac{3}{5}\text{ hrs}$
$\displaystyle \frac{72}{x}-\frac{72}{x+10}=\frac{3}{5}$
$\displaystyle 5[72(x+10)-72x]=3x(x+10)$
$\displaystyle 5[72x+720-72x]=3x(x+10)$
$\displaystyle 3600=3x^2+30x$
$\displaystyle x^2+10x-1200=0$
$\displaystyle (x-30)(x+40)=0$
$\displaystyle x=30\text{ or }x=-40$
$\displaystyle \text{Since speed cannot be negative, }x=30$
$\displaystyle \therefore \text{Original speed of the car }=30\text{ km/hr}$
$\\$

$\displaystyle \textbf{Question 3: } \text{Solve the quadratic equation }x^2-3x-9=0\text{ for }x\text{ and give your} \\ \text{answer correct to two decimal places. (ICSE 2007)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given quadratic equation is }x^2-3x-9=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=-3,c=-9.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-3)\pm\sqrt{(-3)^2-4\times1\times(-9)}}{2\times1}$
$\displaystyle =\frac{3\pm\sqrt{9+36}}{2}=\frac{3\pm\sqrt{45}}{2}$
$\displaystyle =\frac{3\pm6.708}{2}\quad[\because\sqrt{45}=6.708]$
$\displaystyle \text{Taking '+' sign, }x=\frac{3+6.708}{2}=\frac{9.708}{2}=4.854=4.85$
$\displaystyle \text{Taking '-' sign, }x=\frac{3-6.708}{2}=\frac{-3.708}{2}=-1.854=-1.85$
$\displaystyle \text{Hence, the solutions are }4.85\text{ and }-1.85.$
$\displaystyle \text{Hence }x=4.85\text{ or }x=-1.85.$

$\displaystyle \text{Question 4: Five years ago, a woman's age was the square of her son's age. Ten years hence her} \\ \text{age will be twice that of her son's age. Find: (i) the age of the son five years ago. (ii) the present age} \\ \text{of the woman.}\hfill \text{[ICSE 2007]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the age of the son }5\text{ years ago }=x\text{ years}$
$\displaystyle \text{The woman's age }5\text{ years ago }=x^2\text{ years}$
$\displaystyle \text{Present age of the son }=(x+5)\text{ years}$
$\displaystyle \text{Present age of the woman }=(x^2+5)\text{ years}$
$\displaystyle \text{Age of the son }10\text{ years hence }=(x+15)\text{ years}$
$\displaystyle \text{Age of the woman }10\text{ years hence }=(x^2+15)\text{ years}$
$\displaystyle x^2+15=2(x+15)$
$\displaystyle x^2+15=2x+30$
$\displaystyle x^2-2x-15=0$
$\displaystyle (x+3)(x-5)=0$
$\displaystyle x=-3\text{ or }x=5$
$\displaystyle \text{Since age cannot be negative, }x=5$
$\displaystyle \therefore \text{(i) Age of the son }5\text{ years ago }=5\text{ years}$
$\displaystyle \text{(ii) Present age of the woman }=x^2+5=25+5=30\text{ years}$
$\\$

$\displaystyle \textbf{Question 5: } \text{Solve the quadratic equation }5x(x+2)=3\text{ and give your answer correct} \\ \text{to two decimal places. (ICSE 2008)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }5x(x+2)=3\Rightarrow5x^2+10x-3=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get }a=5,b=10,c=-3.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-10\pm\sqrt{(10)^2-4\times5\times(-3)}}{2\times5}$
$\displaystyle =\frac{-10\pm\sqrt{100+60}}{10}=\frac{-10\pm\sqrt{160}}{10}$
$\displaystyle =\frac{-10\pm12.649}{10}\quad[\because\sqrt{160}=12.649]$
$\displaystyle \text{Taking '+' sign, }x=\frac{-10+12.649}{10}=\frac{2.649}{10}=0.2649=0.26\text{ (approx)}$
$\displaystyle \text{Taking '-' sign, }x=\frac{-10-12.649}{10}=\frac{-22.649}{10}=-2.2649=-2.26\text{ (approx)}$
$\displaystyle \text{Hence, the solutions are }0.26\text{ and }-2.26.$

$\displaystyle \text{Question 6: Some students planned a picnic. The budget for the food was Rs. }480.\text{ As} \\ \text{eight of them failed to join the party,} \text{the cost of the food for each member increased} \\ \text{by Rs. }10.\text{ Find how many students went for the picnic.}\hfill \text{[ICSE 2008]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the number of students who planned the picnic }=x$
$\displaystyle \text{Budget for food }=\text{Rs. }480$
$\displaystyle \text{Original share of each student }=\text{Rs. }\frac{480}{x}$
$\displaystyle \text{Number of students who went for the picnic }=(x-8)$
$\displaystyle \text{New share of each student }=\text{Rs. }\frac{480}{x-8}$
$\displaystyle \frac{480}{x-8}-\frac{480}{x}=10$
$\displaystyle \frac{480x-480(x-8)}{x(x-8)}=10$
$\displaystyle \frac{480x-480x+3840}{x(x-8)}=10$
$\displaystyle \frac{3840}{x(x-8)}=10$
$\displaystyle 3840=10x(x-8)$
$\displaystyle 384=x^2-8x$
$\displaystyle x^2-8x-384=0$
$\displaystyle (x-24)(x+16)=0$
$\displaystyle x=24\text{ or }x=-16$
$\displaystyle \text{Since the number of students cannot be negative, }x=24$
$\displaystyle \therefore \text{Number of students who went for the picnic }=x-8=24-8=16$
$\\$

$\displaystyle \textbf{Question 7: } \text{Solve the quadratic equation }4x^2-7x+2=0\text{ and give the answer} \\ \text{correct to two significant figures. (ICSE 2009)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }4x^2-7x+2=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get }a=4,b=-7,c=2.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-7)\pm\sqrt{(-7)^2-4\times4\times2}}{2\times4}$
$\displaystyle =\frac{7\pm\sqrt{49-32}}{8}=\frac{7\pm\sqrt{17}}{8}$
$\displaystyle =\frac{7\pm4.123}{8}\quad[\because\sqrt{17}=4.123]$
$\displaystyle \text{Taking '+' sign, }x=\frac{7+4.123}{8}=\frac{11.123}{8}=1.4$
$\displaystyle \text{Taking '-' sign, }x=\frac{7-4.123}{8}=\frac{2.877}{8}=0.359=0.36\text{ (approx)}$
$\displaystyle \text{Hence, the solutions are }1.4\text{ and }0.36.$

$\displaystyle \textbf{Question 8: } \text{Solve the equation }x-\frac{18}{x}=6.\text{ Give your answer correct} \\ \text{to two significant figures. (ICSE 2011)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }x-\frac{18}{x}=6$
$\displaystyle \Rightarrow x^2-18=6x\Rightarrow x^2-6x-18=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get }a=1,b=-6,c=-18.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(-18)}}{2\times1}$
$\displaystyle =\frac{6\pm\sqrt{36+72}}{2}=\frac{6\pm\sqrt{108}}{2}$
$\displaystyle =\frac{6\pm6\sqrt{3}}{2}=\frac{6\pm2(3\sqrt{3})}{2}$
$\displaystyle =3\pm3\sqrt{3}$
$\displaystyle =3\pm3(1.732)$
$\displaystyle =3+5.196,3-5.196$
$\displaystyle =8.196,-2.196$
$\displaystyle \text{Taking '+' sign, }x=3+5.196=8.196=8.20\text{ (approx)}$
$\displaystyle \text{Taking '-' sign, }x=3-5.196=-2.196=-2.20\text{ (approx)}$
$\displaystyle \text{Hence, the solutions are }8.20\text{ and }-2.20.$

$\displaystyle \text{Question 9: Solve the following equation: }x-\frac{18}{x}=6.$
$\displaystyle \text{Give your answer correct to two significant figures.}\hfill \text{[ICSE 2011]}$
$\displaystyle \text{Answer:}$
$\displaystyle x-\frac{18}{x}=6$
$\displaystyle x^2-18=6x$
$\displaystyle x^2-6x-18=0$
$\displaystyle \text{Comparing }x^2-6x-18=0\text{ with }ax^2+bx+c=0,$
$\displaystyle a=1,\ b=-6,\ c=-18$
$\displaystyle \text{Using }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{6\pm\sqrt{(-6)^2-4(1)(-18)}}{2}$
$\displaystyle x=\frac{6\pm\sqrt{108}}{2}$
$\displaystyle x=\frac{6\pm10.392}{2}$
$\displaystyle x=8.196\text{ or }x=-2.196$
$\displaystyle \therefore \text{Correct to two significant figures, }x=8.2\text{ or }x=-2.2$
$\\$

$\displaystyle \textbf{Question 10: } \text{Solve the equation }5x^2-3x-4=0\text{ and give your answer correct} \\ \text{to three significant figures. (ICSE 2012)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }5x^2-3x-4=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get }a=5,b=-3,c=-4.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-3)\pm\sqrt{(-3)^2-4\times5\times(-4)}}{2\times5}$
$\displaystyle =\frac{3\pm\sqrt{9+80}}{10}=\frac{3\pm\sqrt{89}}{10}$
$\displaystyle =\frac{3\pm9.433}{10}$
$\displaystyle \text{Taking '+' sign, }x=\frac{3+9.433}{10}=\frac{12.433}{10}=1.24$
$\displaystyle \text{Taking '-' sign, }x=\frac{3-9.433}{10}=\frac{-6.433}{10}=-0.643$
$\displaystyle \text{Hence, the solutions are }1.24\text{ and }-0.643.$

$\displaystyle \textbf{Question 11: } \text{Solve and calculate the answer correct to two decimal places. (ICSE 2013)}$
$\displaystyle x^2-5x-10=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }x^2-5x-10=0$
$\displaystyle \text{On comparing above equation with }ax^2+bx+c=0,\text{ we get }a=1,b=-5,c=-10.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-5)\pm\sqrt{(-5)^2-4\times1\times(-10)}}{2\times1}$
$\displaystyle =\frac{5\pm\sqrt{25+40}}{2}=\frac{5\pm\sqrt{65}}{2}$
$\displaystyle =\frac{5\pm8.062}{2}$
$\displaystyle \text{Taking '+' sign, }x=\frac{5+8.062}{2}=\frac{13.062}{2}=6.531=6.53$
$\displaystyle \text{Taking '-' sign, }x=\frac{5-8.062}{2}=\frac{-3.062}{2}=-1.531=-1.53$
$\displaystyle \text{Hence, the solutions are }6.53\text{ and }-1.53.$

$\displaystyle \text{Question 12: A shopkeeper buys a certain number of books for Rs. }960.\text{ If the cost per book was Rs. }8\text{ less,}$
$\displaystyle \text{the number of books that could be bought for Rs. }960\text{ would be }4\text{ more. Taking the}$
$\displaystyle \text{original cost of each book as Rs. }x, \ \text{write an equation in }x\text{ and solve it.}\hfill \text{[ICSE 2013]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Original cost of each book }=\text{Rs. }x$
$\displaystyle \text{Number of books bought for Rs. }960=\frac{960}{x}$
$\displaystyle \text{New cost of each book }=\text{Rs. }(x-8)$
$\displaystyle \text{Number of books bought for Rs. }960=\frac{960}{x-8}$
$\displaystyle \frac{960}{x-8}-\frac{960}{x}=4$
$\displaystyle \frac{960x-960(x-8)}{x(x-8)}=4$
$\displaystyle \frac{960x-960x+7680}{x(x-8)}=4$
$\displaystyle \frac{7680}{x(x-8)}=4$
$\displaystyle 7680=4x(x-8)$
$\displaystyle 1920=x^2-8x$
$\displaystyle x^2-8x-1920=0$
$\displaystyle (x-48)(x+40)=0$
$\displaystyle x=48\text{ or }x=-40$
$\displaystyle \text{Since cost cannot be negative, }x=48$
$\displaystyle \therefore \text{Original cost of each book }=\text{Rs. }48$
$\\$

$\displaystyle \textbf{Question 13: } \text{Solve for }x\text{ using the quadratic formula. Write your answer correct} \\ \text{to two significant figures. (ICSE 2014)}$
$\displaystyle (x-1)^2-3x+4=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }(x-1)^2-3x+4=0$
$\displaystyle \Rightarrow x^2+1-2x-3x+4=0$
$\displaystyle \Rightarrow x^2-5x+5=0$
$\displaystyle \text{On comparing Eq. (i) with }ax^2+bx+c=0,\text{ we get }a=1,b=-5,c=5.$
$\displaystyle \text{Using Sridharacharya's formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-5)\pm\sqrt{(-5)^2-4\times1\times5}}{2\times1}$
$\displaystyle =\frac{5\pm\sqrt{25-20}}{2}=\frac{5\pm\sqrt{5}}{2}$
$\displaystyle =\frac{5\pm2.236}{2}$
$\displaystyle \text{Taking '+' sign, }x=\frac{5+2.236}{2}=\frac{7.236}{2}=3.618=3.6$
$\displaystyle \text{Taking '-' sign, }x=\frac{5-2.236}{2}=\frac{2.764}{2}=1.382=1.4$
$\displaystyle \text{Hence, the values of }x\text{ are }3.6\text{ and }1.4.$

$\displaystyle \text{Question 14: A two digit number is such that the product of its digits is }6.\text{ When }9$
$\displaystyle \text{ is added to this number,} \ \text{the digits interchange their places. Find the number.}\hfill \text{[ICSE 2014]}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the required two digit number be }10x+y$
$\displaystyle \text{Then the number obtained by interchanging the digits }=10y+x$
$\displaystyle xy=6$
$\displaystyle 10x+y+9=10y+x$
$\displaystyle 10x-x+y-10y=-9$
$\displaystyle 9x-9y=-9$
$\displaystyle x-y=-1$
$\displaystyle y=x+1$
$\displaystyle x(x+1)=6$
$\displaystyle x^2+x-6=0$
$\displaystyle (x+3)(x-2)=0$
$\displaystyle x=-3\text{ or }x=2$
$\displaystyle \text{Since }x\text{ is a digit, }x=2$
$\displaystyle y=x+1=2+1=3$
$\displaystyle \therefore \text{Required two digit number }=10x+y=10(2)+3=23$
$\\$

$\displaystyle \textbf{Question 15: } \text{Find the value of }k\text{ for which }x=3\text{ is a solution of the quadratic equation } \\ (k+2)x^2-kx+6=0.\text{ Thus, find the other root of the equation. (ICSE 2015)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }(k+2)x^2-kx+6=0 \qquad \text{...(i)}$
$\displaystyle \text{If }x=3,\text{ then }(k+2)(3)^2-k(3)+6=0$
$\displaystyle \Rightarrow 9(k+2)-3k+6=0$
$\displaystyle \Rightarrow 9k+18-3k+6=0$
$\displaystyle \Rightarrow 6k+24=0$
$\displaystyle \Rightarrow 6k=-24\Rightarrow k=-4$
$\displaystyle \text{Now, putting the value of }k=-4\text{ in Eq. (i), we get}$
$\displaystyle -2x^2+4x+6=0$
$\displaystyle \text{Given, one root }x=3$
$\displaystyle \text{Let other root be }\beta,\text{ then }3+\beta=\frac{-\text{Coefficient of }x}{\text{Coefficient of }x^2}$
$\displaystyle =\frac{-4}{-2}=2$
$\displaystyle \Rightarrow 3+\beta=2\Rightarrow \beta=-1$
$\displaystyle \text{Hence, other root is }-1.$

$\displaystyle \textbf{Question 16: } \text{Solve the quadratic equation }x^2-3(x+3)=0.\text{ Give your answer correct} \\ \text{to two significant figures. (ICSE 2016)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }x^2-3(x+3)-9=0$
$\displaystyle \Rightarrow x^2-3x-9=0$
$\displaystyle \text{By quadratic formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-3)\pm\sqrt{(-3)^2-4\times1\times(-9)}}{2}$
$\displaystyle =\frac{3\pm\sqrt{9+36}}{2}=\frac{3\pm\sqrt{45}}{2}$
$\displaystyle =\frac{3\pm6.71}{2}$
$\displaystyle =\frac{9.71}{2},\frac{-3.71}{2}$
$\displaystyle =4.9,-1.8$

$\displaystyle \textbf{Question 17: } \text{Solve the equation }4x^2-5x-3=0\text{ and give your answer correct} \\ \text{to two decimal places. (ICSE 2017)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }4x^2-5x-3=0$
$\displaystyle \text{By comparing this equation with }ax^2+bx+c=0,\text{ we get }a=4,b=-5,c=-3.$
$\displaystyle \text{Discriminant }(D)=b^2-4ac=(-5)^2-4(4)(-3)$
$\displaystyle =25+48=73$
$\displaystyle \text{Thus, the given equation has real roots.}$
$\displaystyle \text{Now, roots are }\alpha=\frac{-b+\sqrt{D}}{2a}=\frac{-(-5)+\sqrt{73}}{2\times4}$
$\displaystyle =\frac{5+\sqrt{73}}{8}=\frac{5+8.54}{8}=\frac{13.54}{8}=1.69$
$\displaystyle \text{and }\beta=\frac{-b-\sqrt{D}}{2a}=\frac{-(-5)-\sqrt{73}}{2\times4}$
$\displaystyle =\frac{5-8.54}{8}=\frac{-3.54}{8}=-0.44$

$\displaystyle \textbf{Question 18: } \text{Solve the quadratic equation }2x^2-4x-3=0\text{ and give your answer correct} \\ \text{to three significant figures. (ICSE 2017)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }2x^2-4x-3=0 \qquad \text{...(i)}$
$\displaystyle \text{On comparing Eq. (i) with }ax^2+bx+c=0,\text{ we get }a=2,b=-4,c=-3.$
$\displaystyle \text{So, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-4)\pm\sqrt{(-4)^2-4(2)(-3)}}{2(2)}$
$\displaystyle =\frac{4\pm\sqrt{16+24}}{4}=\frac{4\pm\sqrt{40}}{4}$
$\displaystyle =\frac{4\pm4\sqrt{5}}{4}=\frac{4(1\pm\sqrt{5})}{4}$
$\displaystyle =1\pm\sqrt{5}$
$\displaystyle \text{Hence, the values of }x\text{ are }3.236\text{ or }-1.236.$

$\displaystyle \textbf{Question 19: } \text{Solve }x^2+7x=7\text{ and give your answer correct to two decimal places.} \\ \text{(ICSE 2018)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation }x^2+7x=7$
$\displaystyle \Rightarrow x^2+7x-7=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=7,c=-7.$
$\displaystyle \text{Now, }x=\frac{-7\pm\sqrt{(7)^2-4(1)(-7)}}{2(1)}$
$\displaystyle =\frac{-7\pm\sqrt{49+28}}{2}=\frac{-7\pm\sqrt{77}}{2}$
$\displaystyle =\frac{-7\pm8.77}{2}$
$\displaystyle =\frac{1.77}{2},\frac{-15.77}{2}$
$\displaystyle =0.88,-7.88$

$\displaystyle \textbf{Question 20: } \text{Solve for }x\text{ the quadratic equation }x^2-4x-8=0.\text{ Give your answer} \\ \text{correct to three significant figures. (ICSE 2019)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }x^2-4x-8=0$
$\displaystyle \text{which is a quadratic equation of the form }ax^2+bx+c=0.$
$\displaystyle \text{Here, }a=1,b=-4,c=-8.$
$\displaystyle \text{By using quadratic formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{4\pm\sqrt{(-4)^2-4\times1\times(-8)}}{2\times1}$
$\displaystyle =\frac{4\pm\sqrt{16+32}}{2}=\frac{4\pm\sqrt{48}}{2}$
$\displaystyle =\frac{4\pm4\sqrt{3}}{2}=\frac{2(2\pm2\sqrt{3})}{2}$
$\displaystyle =2\pm2\sqrt{3}$
$\displaystyle =2\pm2(1.732)$
$\displaystyle =2+3.464,2-3.464$
$\displaystyle =5.464,-1.464$
$\displaystyle =5.46,-1.46\text{ [correct to three significant figures]}$

$\displaystyle \textbf{Question 21: } \text{The difference of two natural numbers is }7\text{ and their product is }450. \\ \text{Find the numbers. (ICSE 2020)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let two natural numbers be }x\text{ and }y\text{ (such that }x>y).$
$\displaystyle \text{According to the question, }x-y=7\Rightarrow x=7+y \qquad \text{...(i)}$
$\displaystyle \text{and }xy=450$
$\displaystyle \Rightarrow (7+y)y=450 \Rightarrow y^2+7y-450=0$
$\displaystyle \Rightarrow y^2+25y-18y-450=0$
$\displaystyle \Rightarrow y(y+25)-18(y+25)=0$
$\displaystyle \Rightarrow (y-18)(y+25)=0$
$\displaystyle \therefore y=18\ [\because y\neq-25]$
$\displaystyle \therefore x=7+18=25$
$\displaystyle \text{Required natural numbers are }25\text{ and }18.$

$\displaystyle \textbf{Question 22: } \text{Solve the quadratic equation }x^2-7x+3=0.\text{ Give your answer correct} \\ \text{to two decimal places. (ICSE 2020)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }x^2-7x+3=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=1,b=-7,c=3.$
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle x=\frac{7\pm\sqrt{(-7)^2-4\times1\times3}}{2\times1}$
$\displaystyle =\frac{7\pm\sqrt{49-12}}{2}=\frac{7\pm\sqrt{37}}{2}$
$\displaystyle =\frac{7\pm6.08}{2}$
$\displaystyle =\frac{13.08}{2},\frac{0.92}{2}\Rightarrow x=6.54,0.46$

$\displaystyle \textbf{Question 23: } \text{One of the roots of the quadratic equation }x^2-8x+5=0\text{ is }7.3166. \\ \text{ (ICSE Semester I 2022)}$
$\displaystyle \text{The root of the equation correct to }4\text{ significant figures is }$
$\displaystyle \text{(a) }7.3166 \qquad \text{(b) }7.317 \qquad \text{(c) }7.316 \qquad \text{(d) }7.32$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) Given, root of the quadratic equation is }7.3166.$
$\displaystyle \text{Here, the significant figures are }7,3,1,6,6.$
$\displaystyle \text{Now, add }1\text{ with the digit at 4th significant place because the 5th digit is greater than }5.$
$\displaystyle \text{The required number is }7.317.$

$\displaystyle \textbf{Question 24: } \text{The roots of the quadratic equation }4x^2-7x+2=0\text{ are }1.390,0.359. \\ \text{ (ICSE Semester I 2022)}$
$\displaystyle \text{The roots correct to }2\text{ significant figures are }$
$\displaystyle \text{(a) }1.39\text{ and }0.36 \qquad \text{(b) }1.3\text{ and }0.35$
$\displaystyle \text{(c) }1.4\text{ and }0.36 \qquad \text{(d) }1.390\text{ and }0.360$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) In }1.390,$
$\displaystyle \text{since the next digit is }0<5,$
$\displaystyle \text{Hence, }1.390=1.39\text{ (correct to 2 significant figures)}$
$\displaystyle \text{Similarly, in }0.359,$
$\displaystyle \text{since the next digit is }9>5,$
$\displaystyle \text{Hence, }0.359=0.36$

$\displaystyle \textbf{Question 25: } \text{Which of the following quadratic equations has }2\text{ and }3\text{ as its roots?} \\ \text{(ICSE Semester I 2022)}$
$\displaystyle \text{(a) }x^2-5x+6=0 \qquad \text{(b) }x^2+5x+6=0$
$\displaystyle \text{(c) }x^2-5x-6=0 \qquad \text{(d) }x^2+5x-6=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, two roots of quadratic equation as }\alpha=2\text{ and }\beta=3.$
$\displaystyle \therefore\ \text{Quadratic equation will be }x^2-(\alpha+\beta)x+\alpha\beta=0$
$\displaystyle \Rightarrow x^2-(2+3)x+(2)(3)=0 \Rightarrow x^2-5x+6=0$

$\displaystyle \textbf{Question 26: } \text{If }3\text{ is a root of the quadratic equation }x^2-px+3=0,\text{ then }p\text{ is equal to } \\ \text{(ICSE 2023)}$
$\displaystyle \text{(a) }4 \qquad \text{(b) }3 \qquad \text{(c) }5 \qquad \text{(d) }2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) The given quadratic equation is }x^2-px+3=0 \qquad \text{...(i)}$
$\displaystyle \therefore\ 3\text{ is a root of this equation.}$
$\displaystyle \text{So, }x=3\text{ will satisfy this equation.}$
$\displaystyle \text{On putting }x=3\text{ in Eq. (i), we get}$
$\displaystyle 3^2-p(3)+3=0$
$\displaystyle \Rightarrow 9-3p+3=0 \Rightarrow 12-3p=0$
$\displaystyle \Rightarrow -3p=-12 \Rightarrow p=4$

$\displaystyle \textbf{Question 27: } \text{Solve the quadratic equation }x^2+4x-8=0.\text{ Give your answer correct} \\ \text{to one decimal place. (ICSE 2023)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }x^2+4x-8=0\qquad\text{...(i)}$
$\displaystyle \text{On comparing Eq. (i) with }ax^2+bx+c=0,\text{ we get }a=1,b=4,c=-8.$
$\displaystyle \text{Using quadratic formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-4\pm\sqrt{(4)^2-4(1)(-8)}}{2(1)}$
$\displaystyle =\frac{-4\pm\sqrt{16+32}}{2}=\frac{-4\pm\sqrt{48}}{2}$
$\displaystyle =\frac{-4\pm4\sqrt{3}}{2}=-2\pm2\sqrt{3}$
$\displaystyle =-2\pm2(1.73205)$
$\displaystyle \text{Taking '+' sign, }x=-2+3.4641=1.46=1.4$
$\displaystyle \text{Taking '-' sign, }x=-2-3.4641=-5.4641=-5.4$

$\displaystyle \textbf{Question 28: } \text{Solve the quadratic equation }7x^2+2x-2=0.\text{ Give your answer correct} \\ \text{to two decimal places. (ICSE 2023)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, quadratic equation is }7x^2+2x-2=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=7,b=2,c=-2.$
$\displaystyle \text{By quadratic formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-2\pm\sqrt{2^2-4(7)(-2)}}{2(7)}$
$\displaystyle =\frac{-2\pm\sqrt{4+56}}{14}=\frac{-2\pm\sqrt{60}}{14}$
$\displaystyle =\frac{-2\pm2\sqrt{15}}{14}=\frac{-1\pm\sqrt{15}}{7}$
$\displaystyle \text{Taking positive sign, }x=\frac{-1+\sqrt{15}}{7}=0.410$
$\displaystyle \text{Taking negative sign, }x=\frac{-1-\sqrt{15}}{7}=-0.696$

$\displaystyle \textbf{Question 29: } \text{Solve the quadratic equation }2x^2-10x+5=0\text{ and give your answer correct} \\ \text{to three significant figures. (ICSE 2024)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given equation is }2x^2-10x+5=0$
$\displaystyle \text{On comparing with }ax^2+bx+c=0,\text{ we get }a=2,b=-10,c=5.$
$\displaystyle \text{Now, using quadratic formula, }x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
$\displaystyle =\frac{-(-10)\pm\sqrt{(-10)^2-4(2)(5)}}{2(2)}$
$\displaystyle =\frac{10\pm\sqrt{100-40}}{4}=\frac{10\pm\sqrt{60}}{4}$
$\displaystyle =\frac{10\pm2\sqrt{15}}{4}=\frac{5\pm\sqrt{15}}{2}$
$\displaystyle \therefore x=\frac{5+\sqrt{15}}{2}\text{ and }\frac{5-\sqrt{15}}{2}$
$\displaystyle \text{Hence, }x=4.436\text{ and }0.564$