# RD Sharma Solutions Class 10 Quadratic Equations Exercise 8.5

## RD Sharma Solutions Class 10 Chapter 8 Exercise 8.5

### RD Sharma Class 10 Solutions Chapter 8 Ex 8.5 PDF Free Download

#### Exercise 8.5

Q.1: Find the discriminant of the following quadratic equations :

1:  2x2 – 5x + 3 = 0

Soln:  2x2 – 5x + 3 = 0

The given equation is in the form of ax2 + bx + c = 0

Here, a = 2 , b = -5 and c = 3

The discriminant, D = b2 – 4ac

D = (-5)2 –  4 x 2 x 3

D =  25 – 24 = 1

Therefore, the discriminant of the following quadratic equation is 1.

2)  x2 + 2x + 4 =0

Soln:  x2 + 2x + 4 =0

The given equation is in the form of ax2 + bx + c = 0

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

The discriminant is :-

D = (2)2 – 4 x 1 x 4

D = 4 – 16 = – 12

The discriminant of the following quadratic equation is = – 12.

3)   (x -1) (2x -1) = 0

Soln: (x -1) (2x -1) = 0

The provided  equation is (x -1) (2x -1) = 0

By solving it, we get  2x2 – 3x + 1 = 0

Now this equation is in the form of  ax2 + bx + c = 0

Here,  a =  2 , b = -3 , c = 1

The discriminant is :-

D = (-3)2 – 4 x 2 x 1

D =  9 – 8 = 1

The discriminant of the following quadratic equation is = 1.

4)  x2 – 2x + k = 0

Soln:  x2 – 2x + k = 0

The given equation is in the form of ax2 +  bx + c = 0

Here,  a = 1 , b = -2 , and c = k

D = b2 – 4ac

D = (-2)2  –  4(1)(k)

= 4 – 4k

Therefore, the discriminant, D  of the equation is (4 – 4k)

5)   $\sqrt{3}x^{2} + 2\sqrt{2}x – 2\sqrt{3} = 0$

Soln: $\sqrt{3}x^{2} + 2\sqrt{2}x – 2\sqrt{3} = 0$

The given equation is in the form of ax2 +  bx + c = 0

$here a = \sqrt{3} , b = 2\sqrt{2}x and c = -2\sqrt{3}$

The discriminant is,  D = b2 – 4ac

$(2\sqrt{2})^{2} – (4\times \sqrt{3}\times – 2\sqrt{3})$

D  = 8 + 24 = 32

The discriminant, D of the following equation is 32.

6)  x2 – x + 1 = 0

Soln:    x2 – x + 1 = 0

The given equation  is in the form of ax2 + bx + c = 0

Here,  a =1 , b = -1 and  c = 1

The discriminant is D = b2 – 4ac

(-1)2 – 4 x 1 x 1

1 – 4 = – 3

Therefore, The discriminant D of the following  equation is -3.

Q.2: 1)   16x2 =  24x + 1

Soln:  16x2 – 24x – 1 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 16 , b = -24 and c = – 1

Therefore, the discriminant is given as,

D = (-24)2 – 4(16)(-1)

= 576 + 64

= 640

For a quadratic equation to have real roots, D $\geq$  0.

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the following equation are as follows,

$x = \frac{-(-24)\pm \sqrt{640}}{2(16)}$

$x = \frac{24\pm 8\sqrt{10}}{32}$

$x = \frac{3\pm \sqrt{10}}{4}$

The values of x for both the cases will be :

$x = \frac{3+ \sqrt{10}}{4}$ and,

$x = \frac{3-\sqrt{10}}{4}$

2)  x2 + x + 2 = 0

Soln:   x2 + x + 2 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 1, b = 1 and c = 2.

Therefore, the discriminant is given as,

D  = (1)2 –  4(1)(2)

=  1 – 8

= – 7

For a quadratic equation to have real roots, D $\geq$   0.

Here we find that the equation does not satisfy this condition, hence it does not have real roots.

3) $\sqrt{3}x^{2} + 10x – 8\sqrt{3} = 0$

Soln:  $\sqrt{3}x^{2} + 10x – 8\sqrt{3} = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here,  a = $\sqrt{3}$  , b = 10 and c = $-8 \sqrt{3}$

Therefore, the discriminant is given as,

D = (10)2 – 4($\sqrt{3}$  )( $-8 \sqrt{3}$) = 100 + 96 = 196

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-10 \pm \sqrt{196}}{2\sqrt{3}}$

$x = \frac{-10 \pm 14}{2\sqrt{3}}$

$x = \frac{-5 \pm 7}{\sqrt{3}}$

The values of x for both the cases will be :

$x = \frac{-5 + 7}{\sqrt{3}}$

$x = \frac{2}{\sqrt{3}}$  and,

$x = \frac{-5 – 7}{\sqrt{3}}$

$x = -4\sqrt{3}$

4) 3x2 – 2x + 2 = 0

Soln:  3x2 – 2x + 2 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 3 , b = -2 and c = 2.

Therefore, the discriminant is given as,

D = (-2)2 – 4(3)(2)

= 4 – 24 = -20

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation does not satisfy this condition, hence it has no real roots.

5) $2x^{2} – 2\sqrt{6}x + 3 = 0$

Soln:  $2x^{2} – 2\sqrt{6}x + 3 = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 2 , b = $-2\sqrt{6}$ and  c =3.

Therefore, the discriminant is given as,

D = $(-2\sqrt{6})^{2} – 4(2)(3)$

= 24 – 24 = 0

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(2\sqrt{6})\pm 0}{2(2)}$

$x = \frac{-(\sqrt{6})}{2}$

$x = -\sqrt{\frac{3}{2}}$

6)  3a2x2 + 8abx + 4b2 = 0

Soln:  3a2x2 + 8abx + 4b2 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 3a2, b = 8ab and c = 4b2

Therefore, the discriminant is given as,

D = (8ab)2 – 4(3a2)(4b2)

= 64a2b2 – 48a2b= 16a2b2

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-(8ab)\pm \sqrt{16a^{2}b^{2}}}{2(3a^{2})}$

$x = \frac{-(8ab)\pm 4ab}{6a^{2}}$

$x = \frac{-(4b)\pm 2b}{3a}$

The values of x for both the cases will be :

$x = \frac{-(4b)+ 2b}{3a}$

$x = \frac{-(2b)}{3a}$  and

$x = \frac{-(4b)- 2b}{3a}$

$x = \frac{-2b}{a}$

7.)  $3x^{2} + 2\sqrt{5}x – 5 = 0$

Soln.:  $3x^{2} + 2\sqrt{5}x – 5 = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 3,  b = $2\sqrt{5}$  and c = – 5.

Therefore, the discriminant is given as,

D  = $(2\sqrt{5})^{2} – 4(3)(-5)$

= 20 + 60

= 80

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(2\sqrt{5})\pm \sqrt{80}}{2(3)}$

$x = \frac{-(2\sqrt{5})\pm4 \sqrt{5}}{2(3)}$

$x = \frac{-sqrt{5}\pm2 \sqrt{5}}{3}$

The values of x for both the cases will be :

$x = \frac{-sqrt{5}\ + 2 \sqrt{5}}{3}$

$x = \frac{sqrt{5}}{3}$

And,

$x = \frac{-sqrt{5}\ – 2 \sqrt{5}}{3}$

x = – $\sqrt{5}$

8.)  x2 – 2x + 1 = 0

Soln.:  x2 – 2x + 1 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a =1,  b = – 2 and c = 1

Therefore, the discriminant is given as,

D = (-2)2  – 4(1)(1)

=  4 – 4

= 0

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-2)\pm \sqrt{0}}{2(1)}$

x = 2/2

x = 1

Therefore,  the  equation real roots  and its value is 1

9.)  $2x^{2} + 5\sqrt{3}x + 6 = 0$

Soln.:  $2x^{2} + 5\sqrt{3}x + 6 = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here,  a = 2, b = $5\sqrt{3}$  and c = 6.

Therefore, the discriminant is given as,

D = $(5\sqrt{3})^{2}$ –  4(2)(6)

= 75 – 48

= 27

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows

$x = \frac{-(5\sqrt{3})\pm \sqrt{27}}{2(2)}$

$x = \frac{-(5\sqrt{3})\pm 3\sqrt{3}}{4}$

The values of x for both the cases will be :

$x = \frac{-(5\sqrt{3}) + 3\sqrt{3}}{4}$

$x = \frac{-\sqrt{3}}{2}$

And,

$x = \frac{-(5\sqrt{3}) – 3\sqrt{3}}{4}$

$x = -2\sqrt{3}$

10.)  $\sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0$

Soln.:  $\sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = $\sqrt{2}$ , b = 7 , c = $5\sqrt{2}$

Therefore, the discriminant is given as,

D = $(7)^{2} – 4(\sqrt{2})(5\sqrt{2})$

D = 49 – 40

D = 9

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-(7)\pm \sqrt{9}}{2(\sqrt{2})}$

$x = \frac{-7\pm 3}{2(\sqrt{2})}$

The values of x for both the cases will be :

$x = \frac{-7 + 3}{2(\sqrt{2})}$

$x = -\sqrt{2}$

And

$x = \frac{-7 – 3}{2(\sqrt{2})}$

$x = -\frac{5}{\sqrt{2}}$

11.) $2x^{2} – 2\sqrt{2}x + 1 = 0$

Soln.:  $2x^{2} – 2\sqrt{2}x + 1 = 0$

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 2 , b = $– 2\sqrt{2}$ , c = 1

Therefore, the discriminant is given as,

D = $(-2\sqrt{2})^{2} – 4(2)(1)$

= 8 – 8

= 0

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-(-2\sqrt{2})\pm \sqrt{0}}{2(2)}$

$x = \frac{2 \sqrt{2}}{4}$

$x = \frac{1}{\sqrt{2}}$

12.)  3x2 – 5x + 2 = 0

Soln.:  3x2 – 5x + 2 = 0

The given equation can be written in the form of,  ax2 +  bx + c = 0

the discriminant is given by the following equation,  D = b2 – 4ac

here, a = 3, b = -5 and c = 2.

Therefore, the discriminant is given as,

D = (-5)2 – 4(3)(2)

= 25 – 24

= 1

For a quadratic equation to have real roots,  D $\geq$ 0

Here it can be seen that the equation satisfies this condition, hence it has real roots.

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-(-5) \pm \sqrt{1}}{2(3)}$

$x = \frac{5\pm 1}{6}$

The values of x for both the cases will be :

$x = \frac{5 + 1}{6}$

x = 1

And,

$x = \frac{5 – 1}{6}$

x = 2/3

Q.3)  Solve for x : 1.)  $\frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} , x \neq 2 , 4$.

Soln.:  $\frac{x-1}{x-2} + \frac{x-3}{x-4} = 3\frac{1}{3} , x \neq 2 , 4$

The above equation can be solved as follows:

$\frac{(x-1)(x-4) + (x-3)(x-2)}{(x-2)(x-4)} = \frac{10}{3}$

$\frac{x^{2} – 5x + 4 + x^{2} – 5x + 6 }{x^{2} – 6x + 8} = \frac{10}{3}$

6x2 – 30x + 30 = 10x2 – 60x +80

4x2 – 30x + 50 = 0

2x2 – 15x + 25 = 0

The above equation is in the form of ax2 + bx + c = 0

The discriminant is given by the equation, D = b2 – 4ac

Here, a = 2 , b = -15 , c = 25

D = (-15)2 – 4(2)(25)

= 225 – 200

= 25

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-(-15) \pm \sqrt{25}}{2(2)}$

$x = \frac{15 \pm 5}{4}$

The values of x for both the cases will be :

$x = \frac{15 + 5}{4}$

X  =  5

Also,

$x = \frac{15 – 5}{4}$

$x = \frac{ 5}{2}$

2)   $x + \frac{1}{x} = 3 , x \neq 0$

Soln.:  $x + \frac{1}{x} = 3 , x \neq 0$

The above equation can be solved as follows:

$\frac{x^{2} + 1}{x} = 3$

X2  + 1 = 3x

X2 – 3x + 1 = 0

The above equation is in the form of ax2 + bx + c = 0

The discriminant is given by the equation, D = b2 – 4ac

Here, a = 1 , b = – 3 , c = 1

D = (-3)2 – 4(1)(1)

D = 9 – 4

D = 5

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

$x = \frac{-(-3) \pm \sqrt{5}}{2(1)}$

$x = \frac{3 \pm \sqrt{5}}{2}$

The values of x for both the cases will be :

$x = \frac{3 \ + \sqrt{5}}{2}$

And, $x = \frac{3 \ – \sqrt{5}}{2}$

3.)  $\frac{16}{x} – 1 = \frac{15}{x + 1} , x \neq 0 , -1$

Soln. :  $\frac{16}{x} – 1 = \frac{15}{x + 1} , x \neq 0 , -1$

The above equation can be solved as follows:

$\frac{16 – x}{x} = \frac{15}{x + 1}$

(16 – x)(x + 1) = 15x

16x + 16 – x2 – x = 15x

15x + 16 – x2 – 15x = 0

16 – x2 = 0

X2 – 16 = 0

The above equation is in the form of ax2 + bx + c = 0

The discriminant is given by the equation, D = b2 – 4ac

Here, a = 1 , b = 0 , c = -16

D = (0)2 – 4(1)(-16)

D = 64

the roots of an equation can be found out by using,

$x = \frac{-b \pm \sqrt{D}}{2a}$

Therefore, the roots of the equation are given as follows,

$x = \frac{-0 \pm \sqrt{64}}{2(1)}$

$x = \frac{\pm 8}{2}$

$x = \pm 4$

#### Practise This Question

The probability that a teacher will give an unannounced test during any class meeting is 15.If a student is absent twice, the probability that he will miss atleast one test, is