Quadratic Equations - Solved Examples

Exercise 4.4

Exercise 4.2

Steps in "completing the square" method

For a quadratic equation $ax^2 + bx + c = 0$

I. Divide both sides by $a$ resulting into equation such as below
$x^2 + \dfrac{b}{a}x + \dfrac{c}{a} = 0$

II. Move the term $\dfrac{c}{a}$ to the right side of the equation
$x^2 + \dfrac{b}{a}x = -\dfrac{c}{a}$

III. Complete the square on the left hand side of the equation by adding $\dfrac{b^2}{4a^2}$ on both sides
$x^2 + \dfrac{b}{a}x + \dfrac{b^2}{4a^2} = -\dfrac{c}{a} + \dfrac{b^2}{4a^2}$


(i) $2x^2 - 7x + 3 = 0$

Dividing $2x^2 - 7x + 3 = 0$ by 2, we get
$x^2 - \dfrac{7}{2}x + \dfrac{3}{2} = 0$
$⇒ x^2 - \dfrac{7}{2}x = -\dfrac{3}{2}$
To complete the square on left hand side, let us add $\left(\dfrac{\dfrac{7}{2}}{2}\right)^2$ to both sides of the equation
$⇒ x^2 - \dfrac{7}{2}x + \left(\dfrac{7}{4}\right)^2= -\dfrac{3}{2} + \left(\dfrac{7}{4}\right)^2$
$⇒ \left(x - \dfrac{7}{4}\right)^2= -\dfrac{3}{2} + \dfrac{49}{16}$
$⇒ \left(x - \dfrac{7}{4}\right)^2= \dfrac{-24+49}{16}$
$⇒ \left(x - \dfrac{7}{4}\right)^2= \dfrac{25}{16}$
$⇒ x - \dfrac{7}{4}= ±\dfrac{5}{4}$

$∴ x = \dfrac{7}{4} + \dfrac{5}{4}$ or $x = \dfrac{7}{4} - \dfrac{5}{4}$

$⇒ x = \dfrac{12}{4}$ or $x = \dfrac{2}{4}$

$⇒ x = 3$ or $x = \dfrac{1}{2}$

Hence the roots of given quadratic equation are $3$ and $\dfrac{1}{2}$



(ii) $2x^2 + x - 4 = 0$

Dividing $2x^2 + x - 4 = 0$ by 2, we get
$x^2 + \dfrac{1}{2}x - \dfrac{4}{2} = 0$
$⇒ x^2 + \dfrac{1}{2}x = \dfrac{4}{2}$
To complete the square on left hand side, let us add $\left(\dfrac{\dfrac{1}{2}}{2}\right)^2$ to both sides of the equation
$⇒ x^2 + \dfrac{1}{2}x + \left(\dfrac{1}{4}\right)^2= \dfrac{4}{2} + \left(\dfrac{1}{4}\right)^2$
$⇒ \left(x + \dfrac{1}{4}\right)^2= \dfrac{4}{2} + \dfrac{1}{16}$
$⇒ \left(x + \dfrac{1}{4}\right)^2= \dfrac{32+1}{16}$
$⇒ \left(x + \dfrac{1}{4}\right)^2= \dfrac{33}{16}$
$⇒ x + \dfrac{1}{4}= ±\dfrac{\sqrt{33}}{4}$

$∴ x = -\dfrac{1}{4} + \dfrac{\sqrt{33}}{4}$ or $x = -\dfrac{1}{4} - \dfrac{\sqrt{33}}{4}$

$⇒ x = \dfrac{\sqrt{33} - 1}{4}$ or $x = \dfrac{-\sqrt{33} - 1}{4}$

Hence the roots of given quadratic equation are $\dfrac{\sqrt{33} - 1}{4}$ and $\dfrac{-\sqrt{33} - 1}{4}$



(iii) $4x^2 + 4\sqrt{3}x + 3 = 0$

Dividing $4x^2 + 4\sqrt{3}x + 3 = 0$ by 4, we get
$x^2 + \sqrt{3}x + \dfrac{3}{4} = 0$
$⇒ x^2 + \sqrt{3}x = -\dfrac{3}{4}$
To complete the square on left hand side, let us add $\left(\dfrac{\sqrt{3}}{2}\right)^2$ to both sides of the equation
$⇒ x^2 + \sqrt{3}x + \left(\dfrac{\sqrt{3}}{2}\right)^2= -\dfrac{3}{4} + \left(\dfrac{\sqrt{3}}{2}\right)^2$
$⇒ \left(x + \dfrac{\sqrt{3}}{2}\right)^2 = -\dfrac{3}{4} + \dfrac{3}{4}$
$⇒ \left(x + \dfrac{\sqrt{3}}{2}\right)^2 = 0$
$∴ x = -\dfrac{\sqrt{3}}{2}$ or $x = -\dfrac{\sqrt{3}}{2}$

Hence the roots of given quadratic equation are $-\dfrac{\sqrt{3}}{2}$, $-\dfrac{\sqrt{3}}{2}$



(iv) $2x^2 + x + 4 = 0$

Dividing $2x^2 + x + 4 = 0$ by 2, we get
$x^2 + \dfrac{1}{2}x + \dfrac{4}{2} = 0$
$⇒ x^2 + \dfrac{1}{2}x = -\dfrac{4}{2}$
To complete the square on left hand side, let us add $\left(\dfrac{\dfrac{1}{2}}{2}\right)^2$ to both sides of the equation
$⇒ x^2 + \dfrac{1}{2}x + \left(\dfrac{1}{4}\right)^2= -\dfrac{4}{2} + \left(\dfrac{1}{4}\right)^2$
$⇒ \left(x + \dfrac{1}{4}\right)^2= -\dfrac{4}{2} + \dfrac{1}{16}$
$⇒ \left(x + \dfrac{1}{4}\right)^2= \dfrac{-32+1}{16}$
$⇒ \left(x + \dfrac{1}{4}\right)^2= \dfrac{-31}{16}$
$⇒ x + \dfrac{1}{4}= ±\dfrac{\sqrt{-31}}{4}$

As square root of a negative number i.e. $\sqrt{-31}$ is not a real number, real roots do not exist for this quadratic equation.

Find the roots of the quadratic equations given in Q.1 above by applying quadratic formula.

Solution:

For a quadratic equation $ax^2 + bx + c = 0$

if $b^2 - 4ac ≥ 0$, then the roots of the equation are given by $x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$

if $b^2 - 4ac < 0$, then real roots do not exist for the quadratic equation.


(i) $2x^2 - 7x + 3 = 0$

Here $a = 2, b = -7, c = 3$
∴ $b^2 - 4ac$ = $(-7)^2 - 4 × 2 × 3 = 25$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-7) ± \sqrt{25}}{4}$ $ = \dfrac{7 ± 5}{4}$

∴ $x = \dfrac{7 + 5}{4}$ or $x = \dfrac{7 - 5}{4}$
⇒ $x = 3$ or $x = \dfrac{1}{2}$

Hence roots are $3, \dfrac{1}{2}$



(ii) $2x^2 + x - 4 = 0$

Here $a = 2, b = 1, c = -4$
∴ $b^2 - 4ac$ = $(1)^2 - 4 × 2 × -4 = 33$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(1) ± \sqrt{33}}{4}$

∴ $x = \dfrac{-1 + \sqrt{33}}{4}$ or $x = \dfrac{-1 - \sqrt{33}}{4}$

Hence roots are $\dfrac{-1 + \sqrt{33}}{4}$, $\dfrac{-1 - \sqrt{33}}{4}$



(iii) $4x^2 + 4\sqrt{3}x + 3 = 0$

Here $a = 4, b = 4\sqrt{3}, c = 3$
∴ $b^2 - 4ac$ = $(4\sqrt{3})^2 - 4 × 4 × 3 = 0$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-4\sqrt{3} ± \sqrt{0}}{2 × 4}$ $ = \dfrac{-\sqrt{3}}{2}$

⇒ $x = \dfrac{-\sqrt{3}}{2}$ or $x = \dfrac{-\sqrt{3}}{2}$

Hence roots are $\dfrac{-\sqrt{3}}{2}, \dfrac{-\sqrt{3}}{2}$



(iv) $2x^2 + x + 4 = 0$

Here $a = 2, b = 1, c = 4$
∴ $b^2 - 4ac$ = $(1)^2 - 4 × 2 × 4 = -31$

As $b^2 - 4ac$ < 0, no real roots exist for the quadratic equation.

Find the roots of the following equations:

(i) $x - \dfrac{1}{x} = 3$, $x ≠ 0$

(ii) $\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$, $x ≠ -4, 7$

Solution:

Let us covert the given equation to the form $ax^2 + bx + c = 0$

(i) $x - \dfrac{1}{x} = 3$, $x ≠ 0$

⇒ $x^2 - 1 = 3x$ ⇒ $x^2 -3x - 1 = 0$ Here $a = 1, b = -3, c = -1$
∴ $b^2 - 4ac$ = $(-3)^2 - 4 × 1 × -1 = 13$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-3) ± \sqrt{13}}{2 × 1}$ $ = \dfrac{3 ± \sqrt{13}}{2}$

⇒ $x = \dfrac{3 + \sqrt{13}}{2}$ or $x = \dfrac{3 - \sqrt{13}}{2}$

Hence roots are $\dfrac{3 + \sqrt{13}}{2}, \dfrac{3 - \sqrt{13}}{2}$



(ii) $\dfrac{1}{x + 4} - \dfrac{1}{x - 7} = \dfrac{11}{30}$

By cross multiplying, we get

⇒ $\dfrac{(x - 7) - (x + 4)}{(x + 4)(x - 7)} = \dfrac{11}{30}$

⇒ $\dfrac{-11}{x^2 - 3x - 28} = \dfrac{11}{30}$

⇒ $-30 = x^2 - 3x - 28$
⇒ $x^2 - 3x + 2 = 0$

Here $a = 1, b = -3, c = 2$
∴ $b^2 - 4ac$ = $(-3)^2 - 4 × 1 × 2 = 1$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-3) ± \sqrt{1}}{2 × 1}$ $ = \dfrac{3 ± \sqrt{1}}{2}$

⇒ $x = \dfrac{3 + \sqrt{1}}{2}$ or $x = \dfrac{3 - \sqrt{1}}{2}$

⇒ $x = 2$ or $x = 1$

Hence roots are $2, 1$

The sum of the reciprocals of Rehman’s age (in years) 3 years ago and 5 years from now is $\dfrac{1}{3}$. Find his present age.

Solution:

Let the present age of Rehman be $x$
Rehman's age $3$ years ago = $x - 3$
Rehman's age $5$ years from now = $x + 5$

Sum of the reciprocals of Rehman’s age 3 years ago and 5 years from now is $\dfrac{1}{3}$
⇒ $\dfrac{1}{x - 3} + \dfrac{1}{x + 5} = \dfrac{1}{3}$

⇒ $\dfrac{x + 5 + x - 3}{(x - 3)(x + 5)} = \dfrac{1}{3}$

⇒ $\dfrac{2x + 2}{x^2 + 2x - 15} = \dfrac{1}{3}$

⇒ $3(2x + 2) = x^2 + 2x - 15$

⇒ $6x + 6 = x^2 + 2x - 15$
⇒ $x^2 - 4x - 21 = 0$

Here $a = 1, b = -4, c = -21$
∴ $b^2 - 4ac$ = $(-4)^2 - 4 × 1 × (-21) = 100$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-4) ± \sqrt{100}}{2 × 1}$ $ = \dfrac{4 ± 10}{2}$

⇒ $x = \dfrac{4 + 10}{2}$ or $x = \dfrac{4 - 10}{2}$

⇒ $x = 7$ or $x = -3$

As age cannot be negative, Rehman's present age is 7 years.

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.

Solution:

Let the marks scored by Shefali is Mathematics be $x$
Hence, marks scored by her in English will be $30 - x$

2 more marks in Mathematics = $x + 2$
3 less marks in English = $30 - x - 3$ = $27 - x$

∴ $(x + 2)(27 - x) = 210$
⇒ $-x^2 - 2x + 27x + 54 = 210$
⇒ $-x^2 + 25x - 156 = 0$
⇒ $x^2 - 25x + 156 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = -25, c = 156$
∴ $b^2 - 4ac$ = $(-25)^2 - 4 × 1 × 156 = 625 - 624 = 1$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-25) ± \sqrt{1}}{2 × 1}$ $ = \dfrac{25 ± 1}{2}$

⇒ $x = \dfrac{25 + 1}{2}$ or $x = \dfrac{25 - 1}{2}$

⇒ $x = 13$ or $x = 12$

Hence there are 2 possibles answers
If marks scored in Mathematics are $13$, marks scored in English $= 30 - 13 = 17$
If marks scored in Mathematics are $12$, marks scored in English $= 30 - 12 = 18$

The diagonal of a rectangular field is 60 meters more than the shorter side. If the longer side is 30 meters more than the shorter side, find the sides of the fields.

Solution:

Let the shorter side by $x$ meters
Therefore, the longer side will be $(x + 30)$
and, the hypotenuse will be $(x + 60)$

By applying Pythagoras theorem,
$(hypotenusre)^2$ = $(\text{shorter side})^2 + (\text{longer side})^2$
$(x + 60)^2 = x^2 + (x + 30)^2$
$⇒ x^2 + 120x + 3600 = x^2 + x^2 + 60x + 900$
$⇒ x^2 + - 60x - 2700 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = -60, c = -2700$
∴ $b^2 - 4ac$ = $(-60)^2 - 4 × 1 × (-2700) = 3600 + 10800 = 14400$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-60) ± \sqrt{14400}}{2 × 1}$ $ = \dfrac{60 ± 120}{2}$

⇒ $x = \dfrac{60 + 120}{2}$ or $x = \dfrac{60 - 120}{2}$

⇒ $x = 90$ or $x = -30$

As length cannot be negative, $x = 90$

Hence, the length of shorter side = $90$ meters and the length of longer side = $90 + 30 = 120$ meters.

The difference of squares of two numbers is 180. The square of smaller number is 8 times the larger number. Find the two numbers.

Solution:

Let the larger number be $x$
Therefore, the square of smaller number = $8x$

The difference of squares of two numbers is $180$
⇒ $x^2 - 8x = 180$
⇒ $x^2 - 8x - 180 = 0$
Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = -8, c = -180$
∴ $b^2 - 4ac$ = $(-8)^2 - 4 × 1 × (-180) = 64 + 720 = 784$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-8) ± \sqrt{784}}{2 × 1}$ $ = \dfrac{8 ± 28}{2}$

⇒ $x = \dfrac{8 + 28}{2}$ or $x = \dfrac{8 - 28}{2}$

⇒ $x = 18$ or $x = -10$

If the larger number is $-10$, then the square of the smaller number $ = 8 × -10$
As square of a number cannot be negative, larger number cannot be $-10$

Hence, the larger number is $18$
∴ square of the smaller number = $8 × 18 = 144$ and the smaller number is ± 12.

Hence the numbers are $18, 12$ or $18, -12$

A train travels 360 km at a uniform speed. If the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution:

Let the speed of the train be $s$ km/hr and the time taken be $t$ hours.
Distance = Speed × Time = $st$
$∴ 360 = st$
⇒ $t = \dfrac{360}{s}$

If the speed is 5 km/hr more, it would take 1 hour less
⇒ $t - 1 = \dfrac{360}{s + 5}$

⇒ $(t - 1)(s + 5) = 360$
⇒ $st - s + 5t - 5 = 360$

Replacing $t$ by $\dfrac{360}{s}$, we get
⇒ $s × \dfrac{360}{s} - s + 5 × \dfrac{360}{s} - 5 = 360$

⇒ $360 - s + 5 × \dfrac{360}{s} - 5 = 360$

⇒ $-s + \dfrac{1800}{s} - 5 = 0$

⇒ $-s^2 + 1800 - 5s = 0$
⇒ $s^2 + 5s - 1800 = 0$
Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = 5, c = -180$
∴ $b^2 - 4ac$ = $(5)^2 - 4 × 1 × (-1800) = 25 + 7200 = 7225$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(5) ± \sqrt{7225}}{2 × 1}$ $ = \dfrac{-5 ± 85}{2}$

⇒ $x = \dfrac{-5 + 85}{2}$ or $x = \dfrac{-5 - 85}{2}$

⇒ $x = 40$ or $x = -45$

As speed of the train cannot be a negative value, $x = 40$

Hence, speed of the train is $40$ km/hr

Two water taps together can fill a tank in $9\dfrac{3}{8}$ 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.

Solution:

Let the tap of smaller diameter fill the tank in $x$ hours.
Therefore, tap of larger diameter will take $x - 10$ hours to fill the tank.

In $1$ hour, part of the tank filled by tap of smaller diameter = $\dfrac{1}{x}$
In $1$ hour, part of the tank filled by tap of larger diameter = $\dfrac{1}{x - 10}$

In $1$ hour, part of the tank filled by both taps together = $\dfrac{1}{x} + \dfrac{1}{x - 10}$

As two water taps together can fill a tank in $9\dfrac{3}{8}$ hours,
$\dfrac{1}{x} + \dfrac{1}{x - 10}$ = $\dfrac{1}{9\dfrac{3}{8}}$

$\dfrac{1}{x} + \dfrac{1}{x - 10}$ = $\dfrac{8}{75}$

$\dfrac{x - 10 + x}{x(x - 10)}$ = $\dfrac{8}{75}$

$\dfrac{2x - 10}{x^2 - 10x}$ = $\dfrac{8}{75}$

$150x - 750 = 8x^2 - 80x$
$8x^2 - 230x + 750 = 0$
$4x^2 - 115x + 375 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 4, b = -115, c = 375$
∴ $b^2 - 4ac$ = $(-115)^2 - 4 × 4 × 375 = 13225 - 6000 = 7225$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-115) ± \sqrt{7225}}{2 × 4}$ $ = \dfrac{115 ± 85}{8}$

⇒ $x = \dfrac{115 + 85}{8}$ or $x = \dfrac{115 - 85}{8}$

⇒ $x = 25$ or $x = 3.75$

As tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately, $x$ cannot be $3.75$ hours
Hence $x = 25$ hours.

Hence, time taken by the tap of smaller diameter to fill the tank is $25$ hours and that of larger diameter is $25 - 10 = 15$ hours

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 express train is 11 km/hr more than that of passenger train, find the average speed of the two trains.

Solution:

Let the average speed of passenger train be $x$ km/hr
Therefore, the average speed of express train will be $x + 11$ km/hr

Distance = Speed × Time = $st$
$∴ Time = \dfrac{Distance}{Speed}$
Time taken by passenger train to travel $= \dfrac{132}{x}$

Time taken by express train to travel $= \dfrac{132}{x + 11}$

An express train takes 1 hour less than a passenger train to travel 132 km,
$\dfrac{132}{x} - \dfrac{132}{x + 11} = 1$

⇒ $\dfrac{132x + 1452 - 132x}{x(x + 11)} = 1$

⇒ $ 1452 = x^2 + 11x$
⇒ $x^2 + 11x - 1452 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = 11, c = -1452$
∴ $b^2 - 4ac$ = $(11)^2 - 4 × 1 × (-1452) = 121 + 5808 = 5929$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(-11) ± \sqrt{5929}}{2 × 1}$ $ = \dfrac{-11 ± 77}{2}$

⇒ $x = \dfrac{-11 + 77}{2}$ or $x = \dfrac{-11 - 77}{2}$

⇒ $x = 33$ or $x = -44$

As the speed of a train cannot be negative, $x = 33$

Hence speed of passenger train is $33$ km/hr

Speed of express train = $33 + 11 = 44$ km/hr

Sum of the areas of two squares is $468$ $m^2$. If the difference of perimeters is 24 m, find the sides of two squares.

Solution:

Let the side of first square be $x$ and the side of the second square be $y$

Area of a square = $side^2$
∴ $x^2 + y^2 = 468$ ----- (i)

Perimeter of first square = 4 × $side$
$∴ 4x - 4y = 24$
$⇒ 4(x - y) = 24$
$⇒ x - y = 6$
$⇒ x = 6 + y$

Putting this value of $x$ in equation (i), we get
$(6 + y)^2 + y^2 = 468$
$⇒ 36 + 12y + y^2 + y^2 = 468$
$⇒ 2y^2 + 12y - 432 = 0$
$⇒ y^2 + 6y - 216 = 0$

Comparing with equation $ax^2 + bx + c = 0$
$a = 1, b = 6, c = -216$
∴ $b^2 - 4ac$ = $(6)^2 - 4 × 1 × (-216) = 36 + 864 = 900$

∴ the roots are

$x = \dfrac{-b ± \sqrt{b^2 - 4ac}}{2a}$ $ = \dfrac{-(6) ± \sqrt{900}}{2 × 1}$ $ = \dfrac{-6 ± 30}{2}$

⇒ $y = \dfrac{-6 + 30}{2}$ or $y = \dfrac{-6 - 30}{2}$

⇒ $y = 12$ or $y = -18$

As side of a square cannot be negative, $y = 12$

Hence side of first square is $12$ m

Side of second square = $6 + y = 6 + 12 = 18$ m