# Quadratic Equations : Exercise 4.3 (Mathematics NCERT Class 10th)

Q.1      Find the roots of the following quadratic equations, if they exist by the method of completing the square:
(i) $2{x^2} - 7x + 3 = 0$
(ii) $2{x^2} + x - 4 = 0$
(iii) $4{x^2} + 4\sqrt 3 x + 3 = 0$
(iv) $2{x^2} + x + 4 = 0$
Sol.      (i) The equation $2{x^2} - 7x + 3 = 0$ is the same as ${x^2} - {7 \over 2}x + {3 \over 2} = 0$
Now, ${x^2} - {7 \over 2}x + {3 \over 2}$
$= {\left( {x - {7 \over 4}} \right)^2} - {\left( {{7 \over 4}} \right)^2} + {3 \over 2}$
$= {\left( {x - {7 \over 4}} \right)^2} - {{49} \over {16}} + {3 \over 2}$
$= {\left( {x - {7 \over 4}} \right)^2} - {{25} \over {16}}$
Therefore, $2{x^2} - 7x + 3 = 0$
$\Rightarrow$ ${\left( {x - {7 \over 4}} \right)^2} - {{25} \over {16}} = 0$
$\Rightarrow$ ${\left( {x - {7 \over 4}} \right)^2} = {{25} \over {16}}$
$\Rightarrow$ $x - {7 \over 4} = \pm \,{5 \over 4}$
$\Rightarrow$ $x = {7 \over 4} \pm \,{5 \over 4}$
$\Rightarrow$ $x = {7 \over 4} + \,{5 \over 4} = {{12} \over 4} = 3$
$\Rightarrow$ $x = {7 \over 4} - \,{5 \over 4} = {2 \over 4} = {1 \over 2}$
Therefore, The roots of the given equation are 3 and ${1 \over 2}$.

(ii) We have, $2{x^2} + x - 4 = 0$
$\Rightarrow$ ${x^2} + {x \over 2} - 2 = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} - {\left( {{1 \over 4}} \right)^2} - 2 = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} - {1 \over {16}} - 2 = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} - {{33} \over {16}} = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} = {{33} \over {16}}$
$\Rightarrow$ $x + {1 \over 4} = \pm {{\sqrt {33} } \over 4}$
$\Rightarrow$ $x = - {1 \over 4} \pm {{\sqrt {33} } \over 4}$
$\Rightarrow$ $x = {{ - 1 + \sqrt {33} } \over 4}$
$\Rightarrow$ $x = {{ - 1 - \sqrt {33} } \over 4}$
Therefore, The roots of the given equation are ${{ - 1 - \sqrt {33} } \over 4}$ and ${{ - 1 + \sqrt {33} } \over 4}$
(iii) We have, $4{x^2} + 4\sqrt 3 x + 3 = 0$
$\Rightarrow$ ${\left( {2x} \right)^2} + 2 \times \left( {2x} \right) \times \sqrt 3 + {\left( {\sqrt 3 } \right)^2} - {\left( {\sqrt 3 } \right)^2} + 3 = 0$
$\Rightarrow$ ${\left( {2x + \sqrt 3 } \right)^2} - 3 + 3 = 0$
$\Rightarrow$ ${\left( {2x + \sqrt 3 } \right)^2} = 0$
$\Rightarrow$ $x = - {{\sqrt 3 } \over 2}$
$\Rightarrow$ $x = - {{\sqrt 3 } \over 2}$
Therefore, The roots of the given equation are $- {{\sqrt 3 } \over 2}$ and $- {{\sqrt 3 } \over 2}$.

(iv) We have, $2{x^2} + x + 4 = 0$
$\Rightarrow$ ${x^2} + {1 \over 2}x + 2 = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} - {1 \over {16}} + 2 = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} + {{31} \over {16}} = 0$
$\Rightarrow$ ${\left( {x + {1 \over 4}} \right)^2} = - {{31} \over {16}} < 0$
But ${\left( {x + {1 \over 4}} \right)^2}$ cannot be negative for any real value of x. So, there is no real value of x satisfying the given equation. Therefore, the given equation has no real roots.

Q.2      Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
Sol.      (i) The given equation is $2{x^2} - 7x + 3 = 0$
Here a = 2, b = – 7 and c = 3.
Therefore, $D = {b^2} - 4ac = {\left( { - 7} \right)^2} - 4 \times 2 \times 3$ = $49 - 24 = 25 > 0$
So, the given equation has real roots given by
$x = {{ - b \pm \sqrt D } \over {2a}}$
$= {{ - \left( { - 7} \right) \pm \sqrt {25} } \over {2 \times 2}}$
$= {{7 \pm 5} \over 4}$
$= {{12} \over 4}\,or\,\,{2 \over 4} = 3\,\,or\,\,{1 \over 2}$

(ii) The given equation is $2{x^2} + x - 4 = 0$
Here, a = 2, b = 1 and c = – 4
Therefore, $D = {b^2} - 4ac = {\left( 1 \right)^2} - 4 \times 2 \times - 4 = 1 + 32 = 33\, > 0$
So, the given equation has real roots given by
$x = {{ - b \pm \sqrt D } \over {2a}} = {{ - 1 \pm \sqrt {33} } \over {2 \times 2}}$ $= {{ - 1 \pm \sqrt {33} } \over 4}$

(iii) The given equation is $4{x^2} + 4\sqrt 3 x + 3 = 0$
Here a = 4, $b = 4\sqrt 3 \,\,and\,c = 3$
Therefore, $D = {b^2} - 4ac = {\left( {4\sqrt 3 } \right)^2} - 4 \times 4 \times 3 = 48 - 48 = 0$
So, the given equation has real equal roots given by
$x = {{ - b \pm \sqrt D } \over {2a}} = {{ - b} \over {2a}}$ $= {{ - 4\sqrt 3 } \over {2 \times 4}} = {{ - \sqrt 3 } \over 2}$
(iv) The given equation is $2{x^2} + x + 4 = 0$
Here, a = 2, b = 1 and c = 4
Therefore, $D = {b^2} - 4ac = {\left( 1 \right)^2} - 4 \times 2 \times 4 = 1 - 32 = - 31 < 0$
So, the given equation has no real roots.
I prefer to use quadratic formula method as it is a straight forward method.

Q.3      Find the roots of the following equations :
(i) $x - {1 \over x} = 3,x \ne 0$
(ii) ${1 \over {x + 4}} - {1 \over {x - 7}} = {{11} \over {30}},x \ne - 4,7$
Sol.       (i) The given equation is $x - {1 \over x} = 3,x \ne 0$
$\Rightarrow$ ${x^2} - 3x - 1 = 0$
Here , a = 1, b = – 3 and c = – 1
Therefore, $D = {b^2} - 4ac = {\left( { - 3} \right)^2} - 4\left( 1 \right)\left( { - 1} \right) = 9 + 4 = 13 > 0$
So, the given equation has real roots given by
$x = {{ - b \pm \sqrt D } \over {2a}} = {{ - \left( { - 3} \right) \pm \sqrt {13} } \over {2 \times 1}}$ $= {{3 \pm \sqrt {13} } \over 2}$

(ii) The given equation is ${1 \over {x + 4}} - {1 \over {x - 7}} = {{11} \over {30}},x \ne - 4,7$
$\Rightarrow$ ${{\left( {x - 7} \right) - \left( {x + 4} \right)} \over {\left( {x + 4} \right)\left( {x - 7} \right)}} = {{11} \over {30}}$
$\Rightarrow {{x - 7 - x - 4} \over {\left( {x + 4} \right)\left( {x - 7} \right)}} = {{11} \over {30}}$
$\Rightarrow$ ${{ - 11} \over {\left( {x + 4} \right)\left( {x - 7} \right)}} = {{11} \over {30}}$
$\Rightarrow {{ - 1} \over {{x^2} - 7x + 4x - 28}} = {1 \over {30}}$

$\Rightarrow$ ${{ - 1} \over {{x^2} - 3x - 28}} = {1 \over {30}}$

$\Rightarrow$ $- 30 = {x^2} - 3x - 28$
$\Rightarrow$ ${x^2} - 3x + 2 = 0$
$\Rightarrow {x^2} - 2x - x + 2 = 0$

$\Rightarrow$ $\left( {x - 1} \right)\left( {x - 2} \right) = 0$
$\Rightarrow x\left( {x - 2} \right) - 1\left( {x - 2} \right) = 0$
$\Rightarrow$ $x = 1\,\,\,or\,\,\,2$
Thus x = 1 and x = 2 are the roots of the given equation.

Q.4      The sum of the reciprocals of Rehman's ages, (in years) 3 years ago and 5 years from now is ${1 \over 3}$. Find his present age.
Sol.      Let Rehman's present age be x - years.
As per question, we have
${1 \over {x - 3}} + {1 \over {x + 5}} = {1 \over 3}$
$\Rightarrow$ $3\left( {x + 5} \right) + 3\left( {x - 3} \right) = \left( {x - 3} \right)\left( {x + 5} \right)$
$\Rightarrow$ $3x + 15 + 3x - 9 = {x^2} + 2x - 15$
$\Rightarrow$ ${x^2} - 4x - 21 = 0$
$\Rightarrow$ ${x^2} - 7x + 3x - 21 = 0$

$\Rightarrow$ $\left( {x - 7} \right)\left( {x + 3} \right) = 0$
$\Rightarrow x\left( {x - 7} \right) + 3\left( {x - 7} \right) = 0$
$\Rightarrow \left( {x + 3} \right)\left( {x - 7} \right) = 0$
$\Rightarrow$ $x = 7\,\,\,or\,\,\, - 3$
Therefore, $x = 7$ [Since, age can never be negative]
Thus, Rehman's present age is 7 years.

Q.5      In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Sol.      Let Shefali's marks in Mathematics be x. Then her marks in English will be (30 – x).
As per condition of the problem :
$\left( {x + 2} \right) \times \left[ {\left( {30 - x} \right) - 3} \right] = 210$
$\Rightarrow$ $\left( {x + 2} \right)\left( {27 - x} \right) = 210$
$\Rightarrow$ $27x - {x^2} + 54 - 2x = 210$
$\Rightarrow$ ${x^2} - 25x + 156 = 0$
$\Rightarrow$ ${x^2} - 12x - 13x + 156 = 0$
$\Rightarrow$ $x\left( {x - 12} \right) - 13\left( {x - 12} \right) = 0$
$\Rightarrow$ $\left( {x - 12} \right)\left( {x - 13} \right) = 0$
$\Rightarrow$ $x = 12\,\,\,or\,\,\,x = 13$
Therefore, Shafali's marks in Mathematics and English are 12 and 18 respectively or in Mathematics and English are 13 and 17, respectively.

Q.6      The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Sol.      Let in the rectangular field BC = x metres. Then AC = (x + 60) and AB = (x + 30) metres. By Pythagoras Theorem, we have

$A{C^2} = B{C^2} + A{B^2}$
$\Rightarrow$ ${\left( {x + 60} \right)^2} = {x^2} + {\left( {x + 30} \right)^2}$
$\Rightarrow$ ${x^2} + 120x + 3600 = {x^2} + {x^2} + 60x + 900$
$\Rightarrow$ ${x^2} - 60x - 2700 = 0$
$\Rightarrow$ ${x^2} - 90x + 30x - 2700 = 0$
$\Rightarrow$ $x\left( {x - 90} \right) + 30\left( {x - 900} \right) = 0$
$\Rightarrow$ $\left( {x + 30} \right)\left( {x - 90} \right) = 0$
$\Rightarrow$ $x = - 30\,\,\,or\,\,\,x = 90$
$\Rightarrow$ $x = 90\,\,$ [Since side of a rectangle can never be negative]
Hence, the sides of the field are 120 m and 90 m.

Q.7      The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Sol.    Let the larger number be x. Then ,
Square of the smaller number = 8x
Also , square of the larger number = ${x^2}$
It is given that the difference of the squares of the number is 180
Therefore, ${x^2} - 8x = 180$
$\Rightarrow$ ${x^2} - 8x - 180 = 0$
$\Rightarrow$ ${x^2} - 18x + 10x - 180 = 0$
$\Rightarrow$ $x\left( {x - 18} \right) + 10\left( {x - 18} \right) = 0$
$\Rightarrow$ $\left( {x - 18} \right)\left( {x + 10} \right) = 0$
$\Rightarrow$ $x - 18 = 0\,\,\,or\,\,\,x + 10 = 0$
$\Rightarrow$ $x = 18\,\,\,or\,\,\,x = - 10$
Case 1 : When x = 18. In this case, we have
Square of the smaller number = 8x = 8 × 18 = 144
Therefore, Smaller number $= \pm 12$
Thus, the number are 18, 12 or 18, – 12.
Case 2 : When x = – 10
In this case, we have
Square of the smaller number = 8x = 8 × – 10 = – 80
But , square of a number is always positive.
Therefore, x = – 10 is not possible.
Hence, the numbers are 18, 12 or 18 , – 12.

Q.8      A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Sol.       Let x km/hr be the uniform speed of the train.
Then time taken to cover $360\,km = {{360} \over x}hours$.
Time taken to cover 360 km when the speed is increased by $5km/hr = {{360} \over {x + 5}}hours.$
It is given that the time to cover 360 km is reduced by 1 hour.
Therefore, ${{360} \over x} - {{360} \over {x + 5}} = 1$
$\Rightarrow$ $360\left( {x + 5} \right) - 360x = x\left( {x + 5} \right)$
$\Rightarrow$ $360x + 1800 - 360x = {x^2} + 5x$
$\Rightarrow$ ${x^2} + 5x - 1800 = 0$
$\Rightarrow$ ${x^2} + 45x - 40x - 1800 = 0$
$\Rightarrow$ $x\left( {x + 45} \right) - 40\left( {x + 45} \right) = 0$
$\Rightarrow$ $\left( {x - 40} \right)\left( {x + 45} \right) = 0$
$\Rightarrow$ $x - 40 = 0\,\,\,or\,\,\,x + 45 = 0$
$\Rightarrow$ $x = 40\,\,\,or\,\,\,x = - 45$
But x cannot be negative. Therefore, x = 40.
Hence, the original speed of the train is 40 km/hr.

Q.9     Two water taps together can fill a tank in $9{3 \over 8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Sol.        Let the smaller tap takes x hours to fill the tank. So, the larger tap will take (x – 10) hours to fill the tank.
Therefore, Portion of the tank filled by the larger tap in one hours $= {1 \over {x - 10}}$
$\Rightarrow$ Portion of the tank filled by the larger tap $9{3 \over 8}\,hours$
i.e., ${{75} \over 8}\,hours = {1 \over {x - 10}} \times {{75} \over 8}$
Similarly, portion of the tank filld by the smaller tap in ${{75} \over 8}hours = {1 \over x} \times {{75} \over 8}$
Since it is given that the tank is filled in ${{75} \over 8}hours$
Therefore, ${{75} \over {8\left( {x - 10} \right)}} + {{75} \over {8x}} = 1$
$\Rightarrow$ ${1 \over {x - 10}} + {1 \over x} = {8 \over {75}}$
$\Rightarrow$ ${{x + x - 10} \over {x\left( {x - 10} \right)}} = {8 \over {75}}$
$\Rightarrow$ $75\left( {2x - 10} \right) = 8x\left( {x - 10} \right)$
$\Rightarrow$ $150x - 750 = 8{x^2} - 80x$
$\Rightarrow$ $8{x^2} - 230x + 750 = 0$
$\Rightarrow$ $4{x^2} - 115x + 375 = 0$
$\Rightarrow$ $4{x^2} - 100x - 15x + 375 = 0$
$\Rightarrow$ $4x\left( {x - 25} \right) - 15\left( {x - 25} \right) = 0$
$\Rightarrow$ $\left( {x - 25} \right)\left( {4x - 15} \right) = 0$
$\Rightarrow$ $x - 25 = 0\,\,\,or\,\,4x - 15 = 0$
$\Rightarrow$ $x = 25\,\,\,or\,\,{{15} \over 4}$
Therefore, $x = 25\,\,as\,x = \,{{15} \over 4}$ is inadmissible
Hence, the larger tap fills the tank in 15 hours and the smaller tap takes 25 hours to fill the tank.

Q.10      An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, ,find the average speed of the two trains.
Sol.      Let the speed of the express train be x km/hr.
Then, the speed of the passenger train is (x – 11) km/hr
Distance to be covered by train = 132
Time taken by express train $= {{132} \over x}hours$ and, time taken by the passenger train $= {{132} \over {x - 11}}$
By the given condition, ${{132} \over {x - 11}} - {{132} \over x} = 1$
$\Rightarrow$ $132x - 132x + 1452 = {x^2} - 11x$
$\Rightarrow$ ${x^2} - 11x - 1452 = 0$
$\Rightarrow$ ${x^2} - 44x + 33x - 1452 = 0$
$\Rightarrow$ $x\left( {x - 44} \right) + 33\left( {x - 44} \right) = 0$
$\Rightarrow$ $\left( {x - 44} \right)\left( {x + 33} \right) = 0$
$\Rightarrow$ $x = 44\,$
or $x = - 33\,$
But x cannot be negative. Therefore, x = 44
Therefore, Speed of express train = 44 km/hr
and, speed of passenger train = (44 – 11) km/hr = 33 km/hr.

Q.11     Sum of the areas of two squares is $468\,{m^2}$. If the difference of their perimeters is 24 m, find the sides of the two squares.
Sol.       Let the sides of the squares be x and y metres (x > y).
According to question :
${x^2} + {y^2} = 468$ ... (1)
and $4x - 4y = 24$
$\Rightarrow$ $x - y = 6$ ... (2)
Putting x = y + 6 in (1), we get ${\left( {y + 6} \right)^2} + {y^2} = 468$
$\Rightarrow {y^2} + 12y + 36 + {y^2} = 468$
$\Rightarrow$ $2{y^2} + 12y + 36 - 468 = 0$
$\Rightarrow 2{y^2} + 12y - 432 = 0$
$\Rightarrow$ ${y^2} + 6y - 216 = 0$
$\Rightarrow$ ${y^2} + 18y - 12y - 216 = 0$
$\Rightarrow$ $y\left( {y + 18} \right) - 12\left( {y + 18} \right) = 0$
$\Rightarrow$ $\left( {y + 18} \right)\left( {y - 12} \right) = 0$
$\Rightarrow$ $y = - 18\,\,\,\,or\,\,\,y = 12$
But y cannot be negative. Therefore, y = 12
Therefore, x = y + 6 = 12 + 6 = 18
Therefore, The sides of the squares are 18 m and 12 m.