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

Question 1: Car A travels x km for every liter of petrol, while car B travels (x + 5) km for every liter of petrol.
(i) Write down the number of liters of petrol used by car A and car B in covering a distance of 400 km.
(ii) If car A uses 4 liters of petrol more than car B in covering the 400 km, write down an equation in x and solve it to determine the number of liters of petrol used by car B for the journey. [ ICSE Board 1997]
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) No. of liters of petrol used by car A } = \frac{400}{x} \text{liters }$
$\displaystyle \text{No. of liters of petrol used by car A } = \frac{400}{x+5} \text{liters }$
$\displaystyle \text{(ii) Given: } \frac{400}{x} - \frac{400}{x+5} = 4$
$\displaystyle \Rightarrow \frac{400x + 2000 - 400x}{x(x+5)} = 4$
$\displaystyle \Rightarrow 4 ( x^2 + 5x) = 2000$
$\displaystyle \Rightarrow x^2 + 5x - 500 = 0$
$\displaystyle \Rightarrow x = - 25 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 20$
$\displaystyle \text{Therefore No. of litres of petrol used by car B } = \frac{400}{x+5} = \frac{400}{20 + 5} = 16 \text{ liters }$

Question 2: By increasing the speed of a car by 10 km/hr, the time of journey for a distance of 72 km is reduced by 36 minutes. Find the original speed of the car. [ ICSE Board 2005]
$\displaystyle \text{Answer:}$
Let the original speed of the car $= x$ km/hr
$\displaystyle \text{Time taken by it to cover 72 km } = \frac{72}{x}$ hrs
New speed of the car $= (x+10)$ km/hr
$\displaystyle \text{Therefore new time taken by the car to cover 72 km } = \frac{72}{x+ 10}$ hrs
Given : Time is reduced by 36 minutes :
$\displaystyle \Rightarrow \frac{72}{x} - \frac{72}{x+ 10} = \frac{36}{60}$
$\Rightarrow 60[ 72(x+10) -72x] = 36 x ( x+10)$
$\Rightarrow 60[ 72x + 720 - 72x] = 36 x(x+10)$
$\Rightarrow x^2 + 10 x - 1200 = 0$
$\Rightarrow (x-30)(x+40) = 0$
$\Rightarrow x= 30 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = -40$
Since speed cannot be negative hence the value of $x = 30$
The original speed of the car $= 30$ km/hr

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

Question 4: Five years ago, a woman’s age was the square of her son’s age. Ten years hence her age will be twice that of her son’s age. Find :
(i) the age of the son five years ago.
(ii) the present age of the woman [ICSE Board 20007]
$\displaystyle \text{Answer:}$
Let the age of the son 5 years ago $= x$ years
The woman’s age 5 years ago $= x^2$ years
The present age of the woman $= (x^2+5)$ years
The present age of her son $= (x+5)$ years
The woman’s age will be $= ( x^2 + 5) + 10 = ( x^2 + 15)$ years
Her son’s age will be $= (x+5) + 10 = (x+15)$ years
According to the given statement :
10 years hence, woman’s age = twice her son’s age
$\Rightarrow x^2 + 15 = 2 ( x + 15)$
$\Rightarrow x^2 - 2x - 15 = 0$
$\Rightarrow (x+3)(x-5) = 0$
$\Rightarrow x = -3 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 5$
Age cannot be negative so $x = 5$
(i) The age of the son 5 years ago $= x = 5$ years
(i0 The present age of the woman $= (x^2 + 5) = 25 + 5 = 30$ years

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

Question 6: Some students planned a picnic. The budget for the food was Rs. 480. As eight of them failed to join the party, the cost of the food for each member increased by Rs. 10. Find how many students went for the picnic. [ICSE Board 2008]
$\displaystyle \text{Answer:}$
Let the number of students who planned the picnic $= \text{ Rs. } x$
Since, the budget for food for all $= \text{ Rs. } 480$
$\displaystyle \text{Share of each in it }= \text{ Rs. } \frac{480}{x}$
Given eight of them failed to join the party. Therefore $(x-8)$ students went for the picnic
$\displaystyle \text{So, the share of each will be } = \text{ Rs. } \frac{480}{x-8}$
Since, the cost of food for each member (student) is increased by $\text{ Rs. } 10$
$\displaystyle \frac{480}{x-8} - \frac{480}{x} = 10$
$\displaystyle \Rightarrow \frac{480x - 480x + 3340}{x(x-8)}= 10$
$\displaystyle \Rightarrow 3840 = 10 (x^2 - 8x)$
$\displaystyle \Rightarrow x^2 - 8x - 3840 = 0$
On solving, it gives $x = 24 \text{ and } x = - 16$
But the number of students cannot be negative, therefore $x =24$
The number of student who went for picnic $= x-8 = 24-8 = 16$

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

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

$\displaystyle \text{Question 9: Solve the following equation: } x - \frac{18}{x} = 6.$
Give your answer correct to two significant figures. [ICSE Board 2011]
$\displaystyle \text{Answer:}$
$\displaystyle x - \frac{18}{x} = 6 \hspace{1.0cm} \Rightarrow x^2 - 18 = 6x \hspace{1.0cm} \Rightarrow x^2 - 6x - 18 = 0$
$\displaystyle \text{Comparing } x^2 - 6x - 18 = 0 \text{ with } ax^2 + bx + c = 0, \\ \\ \text{ we get: } a = 1, b = -6 \text{ and } c= -18$
$\displaystyle \text{We know } x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Substituting the values we get
$\displaystyle x = \frac{6 \pm \sqrt{(6)^2 - 4(1)(-18)}}{2(1)} = \frac{6 \pm 10.392}{2}$
$\displaystyle \Rightarrow x = 8.196 \text{ and } x = -2.196$
or correct to 2 decimal places: $\displaystyle x = 8.2 \text{ and } x = -2.2$

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

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

Question 12: A shopkeeper buys a certain number of boots for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be bought for Rs. 960 would be 4 more. Taking the original cost of each book to be Rs. x, write and equation in x and solve it. [ICSE Board 2013]
$\displaystyle \text{Answer:}$
Original cost of each book $= \text{ Rs. } x$
$\displaystyle \text{No. of books bought for} \text{ Rs. } 960 = \frac{960}{x}$
In 2nd Case:
The cost of each book $= \text{ Rs. } (x-8)$
$\displaystyle \text{No. of books bought for} \text{ Rs. } 960 =\frac{960}{x-8}$
$\displaystyle \text{Given: } \frac{960}{x-8} - \frac{960}{x} = 4$
$\displaystyle \Rightarrow \frac{960x- 960x+7680}{x(x-8)} = 4$
$\Rightarrow 7680 = 4 ( 4x^2 - 8x)$
$\Rightarrow x^2 - 8x = 1920$
$\Rightarrow x^2 - 8x - 1920 = 0$
$\Rightarrow (x-48)(x+40) = 0$
$\Rightarrow x = 48 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = -49$
Since cost cannot be negative, therefore $x = 48$

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

Question 14: A two digit number is such that the product of its digits is 6. When 9 is added to this number; the digits interchange their places. Find the number. [ICSE Board 2014]
$\displaystyle \text{Answer:}$
Let the required two digit number be $10x + y$
$\text{Given: } xy = 6 \text{ and } 10x + y + 9 = 10y + x$
$\Rightarrow 10x+ y + 9 = 10y +x$
$\Rightarrow 9y =9x + 9$
$\Rightarrow y = x + 1$
Now, $xy = 6$
$\Rightarrow x(x+1)=6$
$\Rightarrow x^2 + x - 6 = 0$
$\Rightarrow (x+3)(x-2) = 0$
$x = -3 \hspace{0.5cm} \text{ or } \hspace{0.5cm} x = 2$
Since, $-3$ is not a digit, $x = 2$
$y = x+1 = 2+1 = 3$
Therefore The required two digit number $= 10x + y = 10(2) + 3 = 23$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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