CBSE 10th Maths - Quadratic-Equations - Exercise 4.3

Solution:

(i) $2x^{2} – 7x + 3$ = 0

$⇒ 2x^{2} – 7x$ = – 3

Dividing by 2, on both sides, we get,

⇒ $x^{2} -\dfrac{7x}{2} $ = $\dfrac{-3}{2} $

⇒ $x^{2} -2 × x ×\dfrac{7}{4} $ = $\dfrac{-3}{2} $

On adding $(\dfrac{7}{4} )^{2} $ to both sides of the equation, we get,

⇒ $(x)^{2} -2×x×\dfrac{7}{4} +(\dfrac{7}{4})^{2} $ = $(\dfrac{7}{4})^{2} -\dfrac{3}{2} $

⇒ $(x-\dfrac{7}{4})^{2}$ = $(\dfrac{49}{16}) – (\dfrac{3}{2})$

⇒$(x-\dfrac{7}{4})^{2}$ = $\dfrac{25}{16}$

⇒$(x-\dfrac{7}{4})^{2}$ = $±\dfrac{5}{4}$

⇒ 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 = $\dfrac{1}{2}$

(ii) $2x^{2} + x – 4$ = 0

⇒ $2x^{2} + x $= 4

Dividing both sides of the equation by 2, we get,

⇒ $x^{2} +\dfrac{x}{2}$ = 2

Now on adding$ (\dfrac{1}{4})^{2} $ to both sides of the equation, we get,

⇒ $(x)^{2} + 2 × x × \dfrac{1}{4} + (\dfrac{1}{4})^{2}$ = $2 + (\dfrac{1}{4})^{2} $

⇒$ (x + \dfrac{1}{4})^{2} $ =$ \dfrac{33}{16}$

⇒ $x + \dfrac{1}{4} $= $± \dfrac{\sqrt{33}}{4}$

⇒ x = $± \dfrac{\sqrt{33}}{4} – \dfrac{1}{4}$

⇒ x =$\dfrac{ (± \sqrt{33}-1)}{4}$

Therefore, either x =$\dfrac{ (\sqrt{33}-1)}{4}$ or x = $\dfrac{(-\sqrt{33}-1)}{4}$

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

Converting the equation into $a^{2} +2ab+b^{2} $ form, we get,

⇒ $(2x)^{2} + 2 × 2x × \sqrt{3} + (\sqrt{3})^{2}$ = 0

⇒ $(2x + \sqrt{3})^{2}$ = 0

⇒ $(2x + \sqrt{3}$) = 0 and $(2x + \sqrt{3})$ = 0

Therefore, either x = $\dfrac{-\sqrt{3}}{2}$ or x = $\dfrac{-\sqrt{3}}{2}$.

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

⇒ $2x^{2} + x $= -4

Dividing both sides of the equation by 2, we get

⇒ $x^{2} + \dfrac{1}{2x} $= 2

⇒$x^{2} + 2 × x × \dfrac{1}{4}$ = -2

By adding ( $\dfrac{1}{4} )^{2}$ to both sides of the equation, we get

⇒ $(x)^{2} + 2 × x × \dfrac{1}{4} + ( \dfrac{1}{4} )^{2} $ = ( $\dfrac{1}{4} )^{2} $ – 2

⇒ (x + $\dfrac{1}{4} )^{2}$ = $\dfrac{1}{16} $– 2

⇒ (x + $\dfrac{1}{4} )^{2}$ = - $\dfrac{31}{16} $

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, $2x^{2} + x + 4$ = 0.

Solution:

(i) $2x^{2}$ – 7x + 3 = 0

On comparing the given equation with $ax^{2} + bx + c $= 0, we get,

a = 2, b = -7 and c = 3

By using the quadratic formula, we get,

x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

⇒ x = $\dfrac{(7\pm\sqrt{(49 – 24)})}{4}$

⇒ x = $\dfrac{(7±\sqrt{25})}{4}$

⇒ x = $\dfrac{(7±5)}{4}$

⇒ x = $\dfrac{(7+5)}{4}$or x = $\dfrac{(7-5)}{4}$

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

∴ x = 3 or $\dfrac{1}{2}$

(ii) 2x2 + x – 4 = 0

On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,

a = 2, b = 1 and c = -4

By using the quadratic formula, we get,

x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

⇒x = $\dfrac{(-1±\sqrt{1+32})}{4}$

⇒x = $\dfrac{(-1±\sqrt{33})}{4}$

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

(iii) 4x2 + $4\sqrt{3}x$ + 3 = 0

On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,

a = 4, b = $4\sqrt{3}$ and c = 3

By using the quadratic formula, we get,

x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

⇒ x = $\dfrac{(-4\sqrt{3}±\sqrt{48-48})}{8}$

⇒ x = $\dfrac{(-4\sqrt{3}±0)}{8}$

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

(iv) 2x2 + x + 4 = 0

On comparing the given equation with $ax^{2} + bx + c$ = 0, we get,

a = 2, b = 1 and c = 4

By using the quadratic formula, we get

x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

⇒ x = $\dfrac{(-1±\sqrt{1-32})}{4}$

⇒ x = $\dfrac{(-1±\sqrt{-31})}{4}$

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

Solution:

(i) $x-\dfrac{1}{x}$ = 3

⇒ $x^{2} – 3x -1$ = 0

On comparing the given equation with $ax^{2}+ bx + c $= 0, we get

a = 1, b = -3 and c = -1

By using the quadratic formula, we get

x=$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

⇒ x = $\dfrac{(3±\sqrt{9+4})}{2}$

⇒ x = $\dfrac{(3±\sqrt{13})}{2}$

∴ x = $\dfrac{(3+\sqrt{13})}{2}$ or x = $\dfrac{(3-\sqrt{13})}{2}$

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

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

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

⇒ (x+4)(x-7) = -30

⇒ $x^{2} – 3x – 28 $= 30

⇒$ x^{2} – 3x + 2 $= 0

We can solve this equation by factorisation method now,

⇒ $x^{2} – 2x – x + 2 $= 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

Solution:

Let us say the present age of Rehman is x years.

Three years ago, Rehman’s age was (x – 3) years.

Five years after, his age will be (x + 5) years.

Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to $\dfrac{1}{3}$.

∴ $ \dfrac{1}{x}-3 + \dfrac{1}{x}-5$ = $\dfrac{1}{3}$

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

$\dfrac{(2x+2)}{(x-3)(x+5)}$ =$\dfrac{1}{3}$

⇒ 3(2x + 2) = (x-3)(x+5)

⇒ 6x + 6 = $x^{2} + 2x – 15$

⇒ $x^{2} – 4x – 21$ = 0

⇒ $x^{2} – 7x + 3x – 21$ = 0

⇒ x(x – 7) + 3(x – 7) = 0

⇒ (x – 7)(x + 3) = 0

⇒ x = 7, -3

As we know, age cannot be negative.

Therefore, Rehman’s present age is 7 years.

Solution:

Let us say the marks of Shefali in Maths be x.

Then, the marks in English will be 30 – x.

As per the given question,

(x + 2)(30 – x – 3) = 210

(x + 2)(27 – x) = 210

⇒ $-x^{2} + 25x + 54$ = 210

⇒ $x^{2} – 25x + 156$ = 0

⇒$ x^{2} – 12x – 13x + 156$ = 0

⇒ x(x – 12) -13(x – 12) = 0

⇒ (x – 12)(x – 13) = 0

⇒ x = 12, 13

Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18, and if the marks in Maths are 13, then marks in English will be 30 – 13 = 17.

Solution:

Let us say the shorter side of the rectangle is x m.

Then, larger side of the rectangle = (x + 30) m

Diagonal of the rectangle =$\sqrt{x^{2}+(x+30)^{2}}$

As given, the length of the diagonal is = x + 30 m

Therefore,

$\sqrt{x^{2}+(x+30)^{2}}$=x+60

⇒ $x^{2}2 + (x + 30)^{2}$ = $(x + 60)^{2}$

⇒ $x^{2}+ x^{2} + 900 + 60x $= $x^{2} + 3600 + 120x$

⇒$ x^{2} – 60x – 2700$ = 0

⇒ $x^{2} – 90x + 30x – 2700$ = 0

⇒ x(x – 90) + 30(x -90) = 0

⇒ (x – 90)(x + 30) = 0

⇒ x = 90, -30

However, the side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m, and the length of the larger side will be (90 + 30) m = 120 m.

Solution:

Let us say the larger and smaller number be x and y, respectively.

As per the question given,

$x^{2} – y^{2} $= 180 and $y^{2}$ = 8x

⇒ $x^{2} – 8x$ = 180

⇒ $x^{2} – 8x – 180$ = 0

⇒ $x^{2} – 18x + 10x – 180$ = 0

⇒ x(x – 18) +10(x – 18) = 0

⇒ (x – 18)(x + 10) = 0

⇒ x = 18, -10

The larger number will be 18 only.

x = 18

∴ $y^{2}$ = 8x = 8 × 18 = 144

⇒ y =$ \pm\sqrt{144}$ = ±12

∴ Smaller number = ±12

Therefore, the numbers are 18 and 12 or 18 and -12.

Solution:

It is given that

Distance = 360 km

Consider x as the speed, then the time taken

t = $\dfrac{360}{x}$

If the speed is increased by 5 km/h speed will be (x + 5) km/h

Distance will be the same

t = $\dfrac{360}{x+5}$

We know that

Time with original speed – Time with increased speed = 1

$\dfrac{360}{x}$ – $\dfrac{360}{x+5}$= 1

LCM = x (x + 5)

$\dfrac{[360 (x + 5) – 360x]}{x(x + 5)}$ = 1 360 x + 1800 – 360x = x (x + 5)

$x^{2} + 5x $ = 1800

$x^{2} + 5x – 1800 $ = 0

$x^{2}+ 45x – 40x – 1800 $= 0

x (x + 45) – 40 (x + 45) = 0

(x – 40) (x + 45) = 0

x = 40 km/hr

As we know, the value of speed cannot be negative.

Therefore, the speed of the train is 40 km/h.

Solution:

Let the time taken by the smaller pipe to fill the tank = x hr.

Time taken by the larger pipe = (x – 10) hr

Part of the tank filled by smaller pipe in 1 hour = $\dfrac{1}{x}$

Part of the tank filled by larger pipe in 1 hour =$\dfrac{1}{x-10}$

As given, the tank can be filled in

$9\dfrac{3}{8}$ = $\dfrac{75}{8}$ hours by both the pipes together.

Therefore,

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

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

⇒ $\dfrac{2x-10}{x(x-10)}$ = $\dfrac{8}{75}$

⇒ 75(2x – 10) = $8x^{2} – 80x$

⇒ 150x – 750 =$ 8x^{2} – 80x$

⇒ $8x^{2} – 230x +750 $= 0

⇒ $8x^{2}– 200x – 30x + 750 $= 0

⇒ 8x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(8x -30) = 0

⇒ x = 25, $\dfrac{30}{8}$

Time taken by the smaller pipe cannot be $\dfrac{30}{8}$= 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.

Therefore, the time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours, respectively.

Solution:

Let us say, the average speed of the passenger train = x km/h.

Average speed of express train = (x + 11) km/h

Given, the time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore,

($\dfrac{132}{x}$) – ($\dfrac{132}{(x+11)}$) = 1

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

$\dfrac{132 × 11 }{(x(x+11))}$ = 1

⇒ 132 × 11 = x(x + 11)

⇒ $x^{2} + 11x – 1452 $= 0

⇒ $x^{2} + 44x -33x -1452$ = 0

⇒ x(x + 44) -33(x + 44) = 0

⇒ (x + 44)(x – 33) = 0

⇒ x = – 44, 33

As we know, speed cannot be negative.

Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.