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 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,

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

\( x = \frac{\pm 8}{2} \)

\( x = \pm 4 \)