#### Exercise 8.5

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

1:Â 2x^{2} – 5x + 3 = 0

Soln:Â 2x^{2} – 5x + 3 = 0

The given equation is in the form of ax^{2} + bx + c = 0

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

The discriminant, D = b^{2} – 4ac

D = (-5)^{2} â€“ Â 4 x 2 x 3

D =Â 25 â€“ 24 = 1

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

2)Â x^{2} + 2x + 4 =0

Soln:Â x^{2} + 2x + 4 =0

The given equation is in the form of ax^{2} + bx + c = 0

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

The discriminant is :-

D = (2)^{2} – 4 x 1 x 4

D = 4 â€“ 16 = – 12

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

3)Â Â (x -1) (2x -1) = 0

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

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

By solving it, we get Â 2x^{2} – 3x + 1 = 0

Now this equation is in the form ofÂ ax^{2} + bx + c = 0

Here,Â a = Â 2 , b = -3 , c = 1

The discriminant is :-

D = (-3)^{2} â€“ 4 x 2 x 1

D = Â 9 â€“ 8 = 1

The discriminant of the following quadratic equation is = 1.

4)Â x^{2 }â€“ 2x + k = 0

Soln: Â x^{2 }â€“ 2x + k = 0

The given equation is in the form of ax^{2} +Â bx + c = 0

Here, Â a = 1 , b = -2 , and c = k

D = b^{2} â€“ 4ac

D = (-2)^{2} Â â€“ Â 4(1)(k)

= 4 â€“ 4k

Therefore, the discriminant, DÂ of the equation is (4 â€“ 4k)

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

Soln:Â \( \sqrt{3}x^{2} + 2\sqrt{2}x – 2\sqrt{3} = 0 \)

The given equation is in the form of ax^{2} +Â bx + c = 0

\( here a = \sqrt{3} , b = 2\sqrt{2}x Â and Â c = -2\sqrt{3} \)

The discriminant is,Â D = b^{2} – 4ac

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

DÂ = 8 + 24 = 32

The discriminant, D of the following equation is 32.

Â

6) Â x^{2} â€“ x + 1 = 0

Soln:Â Â Â x^{2} â€“ x + 1 = 0

The given equationÂ is in the form of ax^{2} + bx + c = 0

Here, Â a =1 , b = -1 and Â c = 1

The discriminant is D = b^{2} â€“ 4ac

(-1)^{2} â€“ 4 x 1 x 1

1 – 4 = – 3

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

Â

Q.2:Â 1)Â Â 16x^{2} = Â 24x + 1

Soln: Â 16x^{2} – 24x – 1 = 0

The given equation can be written in the form of, Â ax^{2 }+ Â bx + c = 0

the discriminant is given by the following equation, Â D = b^{2 }– 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 \( \geq \)

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 :

\( x = \frac{3+ \sqrt{10}}{4} \)

\( x = \frac{3-\sqrt{10}}{4} \)

2) Â x^{2} + x + 2 = 0

Soln:Â Â x^{2} + x + 2 = 0

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation, Â D = b^{2} – 4ac

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

Therefore, the discriminant is given as,

D Â = (1)^{2} – Â 4(1)(2)

= Â 1 – 8

= – 7

For a quadratic equation to have real roots, D \( \geq \)

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

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

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

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, Â a = \( \sqrt{3} \)

Therefore, the discriminant is given as,

D = (10)^{2} – 4(\( \sqrt{3} \)

For a quadratic equation to have real roots, Â D \( \geq \)

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 :

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

\( x = \frac{2}{\sqrt{3}} \)

\( x = \frac{-5 – 7}{\sqrt{3}} \)

\( x = -4\sqrt{3} \)

4) 3x^{2} – 2x + 2 = 0

Soln: Â 3x^{2} – 2x + 2 = 0

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 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,Â D \( \geq \)

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:Â \( 2x^{2} – 2\sqrt{6}x + 3 = 0 \)

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a = 2 , b = \( -2\sqrt{6} \)

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,Â D \( \geq \)

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) Â 3a^{2}x^{2 }+ 8abx + 4b^{2} = 0

Soln: Â 3a^{2}x^{2 }+ 8abx + 4b^{2} = 0

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a = 3a^{2}, b = 8ab and c = 4b^{2}

Therefore, the discriminant is given as,

D = (8ab)^{2} – 4(3a^{2})(4b^{2})

= 64a^{2}b^{2} – 48a^{2}b^{2Â }= 16a^{2}b^{2}

For a quadratic equation to have real roots,Â D \( \geq \)

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 :

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

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

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

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

7.) Â \( 3x^{2} + 2\sqrt{5}x – 5 = 0 \)

Soln.: Â \( 3x^{2} + 2\sqrt{5}x – 5 = 0 \)

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a = 3,Â b = \( 2\sqrt{5} \)

Therefore, the discriminant is given as,

D Â = \( (2\sqrt{5})^{2} – 4(3)(-5) \)

= 20 + 60

= 80

For a quadratic equation to have real roots,Â D \( \geq \)

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.) Â x^{2} – 2x + 1 = 0

Soln.:Â x^{2} – 2x + 1 = 0

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a =1,Â b = – 2 and c = 1

Therefore, the discriminant is given as,

D = (-2)^{2} Â – 4(1)(1)

= Â 4 – 4

= 0

For a quadratic equation to have real roots,Â D \( \geq \)

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, Â theÂ equation real roots Â and its value is 1

9.)Â \( 2x^{2} + 5\sqrt{3}x + 6 = 0 \)

Soln.: Â \( 2x^{2} + 5\sqrt{3}x + 6 = 0 \)

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, Â a = 2, b = \(5\sqrt{3}\)

Therefore, the discriminant is given as,

D = \( (5\sqrt{3})^{2} \)

= 75 – 48

= 27

For a quadratic equation to have real roots,Â D \( \geq \)

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.)Â \( \sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0 \)

Soln.:Â \( \sqrt{2}x^{2} + 7x + 5\sqrt{2} = 0 \)

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a = \( \sqrt{2} \)

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,Â D \( \geq \)

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.: Â \( 2x^{2} – 2\sqrt{2}x + 1 = 0 \)

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 4ac

here, a = 2 , b = \( – 2\sqrt{2} \)

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,Â D \( \geq \)

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,

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

\( x = \frac{2 \sqrt{2}}{4} \)

\( x = \frac{1}{\sqrt{2}} \)

Â

12.)Â 3x^{2} – 5x + 2 = 0

Soln.:Â 3x^{2} – 5x + 2 = 0

The given equation can be written in the form of,Â ax^{2 }+Â bx + c = 0

the discriminant is given by the following equation,Â D = b^{2 }– 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,Â D \( \geq \)

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,

\( x = \frac{5 – Â 1}{6} \)

x = 2/3

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

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

The above equation can be solved as follows:

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

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

6x^{2} â€“ 30x + 30 = 10x^{2} â€“ 60x +80

4x^{2} â€“ 30x + 50 = 0

2x^{2} â€“ 15x + 25 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }â€“ 4ac

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

D = (-15)^{2} â€“ 4(2)(25)

= 225 â€“ 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 :

\( x = \frac{15 + 5}{4} \)

XÂ =Â 5

Also,

\( x = \frac{15 – 5}{4} \)

\( x = \frac{ 5}{2} \)

2)Â Â \( x + \frac{1}{x} = 3 , x \neq 0 \)

Soln.:Â \( x + \frac{1}{x} = 3 , x \neq 0 \)

The above equation can be solved as follows:

\( \frac{x^{2} + 1}{x} = 3 \)

X^{2}Â + 1 = 3x

X^{2} â€“ 3x + 1 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }â€“ 4ac

Here, a = 1 , b = – 3 , c = 1

D = (-3)^{2 }– 4(1)(1)

D = 9 â€“ 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 :

\( x = \frac{3 \ + Â \sqrt{5}}{2} \)

And, \( x = \frac{3 \ – Â \sqrt{5}}{2} \)

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

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

The above equation can be solved as follows:

\( \frac{16 – x}{x} = \frac{15}{x + 1} \)

(16 – x)(x + 1) = 15x

16x + 16 â€“ x^{2} â€“ x = 15x

15x + 16 â€“ x^{2} â€“ 15x = 0

16 â€“ x^{2} = 0

X^{2} â€“ 16 = 0

The above equation is in the form of ax^{2} + bx + c = 0

The discriminant is given by the equation, D = b^{2 }â€“ 4ac

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

D = (0)^{2} â€“ 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,

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

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

\( x = \pm 4 \)