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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.5
  • Last Updated : 30 Apr, 2021

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

(i) 2x2 – 5x + 3 = 0  

Solution:

Given quadratic equation: 2x2 – 5x + 3 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get



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

Now, we find the discriminant(D) = b2 – 4ac

D = (-5)2 – 4(2)(3)

= 25 – 24

= 1

Hence, the discriminant of given quadratic equation is 1

(ii) x2 + 2x + 4 = 0

Solution:

Given quadratic equation: x2 + 2x + 4 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get

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

Now, we find the discriminant(D) = b2 – 4ac

D = (2)2 – 4(1)(4)

= 4 – 16

= -12

Hence, the discriminant of given quadratic equation is -12

(iii) (x – 1) (2x – 1)

Solution:



Given quadratic equation:(x – 1)(2x – 1)

Or we can also write as, 2x2 – 3x + 1 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get

Here, a = 2, b = -3 and c = 1

Now, we find the discriminant(D) = b2 – 4ac

D = (-3)2 – 4(2)(1)

= 9 – 8

= 1

Hence, the discriminant of given quadratic equation is 1

(iv) x2 – 2x + k = 0

Solution:

Given quadratic equation: x2 – 2x + k = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get

Here, a = 1, b = -2 and c = k

Now, we find the discriminant(D) = b2 – 4ac

D = (-2)2 – 4(1)(k)

= 4 -4k



Hence, the discriminant of given quadratic equation is 4 – 4k

(v) √3x2 + 2√2x – 2√3 = 0 

Solution:

Given quadratic equation:√3x2 + 2√2x – 2√3 = 0    ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get

Here, a =√3, b = 2√2, and c = -2 – 2√3

Now, we find the discriminant(D) = b2 – 4ac

D = (2√2)2 – 4√3(-2√3)

= 8 + 24

= 32

Hence, the discriminant of given quadratic equation is 32

(vi) x2 – x + 1 = 0

Solution:

Given quadratic equation: x2 – x + 1 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0   ….(2)

On comparing eq(1) and (2), we get

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

Now, we find the discriminant(D) = b2 – 4ac

D = (-1)2 – 4(1)(1)



= 1 – 4

= -3

Hence, the discriminant of given quadratic equation is -3

Question 2. In the following, determine whether the given quadratic equation have real roots and If so, find the roots:

(i) 16x2 = 24x + 1

Solution:

Given quadratic equation: 16x2 – 24x – 1 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 16, b = -24, and c = -1

Now, we find the discriminant(D) = b2 – 4ac

D = (-24)2 – 4(16)(-1)

= 576 + 64

= 640

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

x=\frac{24\pm8\sqrt{1032}}{32}



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

Hence, the value of x is 

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

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

(ii) x2 + x + 2 = 0

Solution:

Given quadratic equation: x2 + x + 2 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (1)2 – 4(1)(2)

= 1 – 8

= -7

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation does not satisfies the given condition, so it does not have real roots.

(iii) √3x2 + 10x – 8√3 = 0

Solution:

Given quadratic equation: √3x2 + 10x – 8√3 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)



On comparing eq(1) and (2), we get  

Here, a = √3, b = 10 and c = -8√3 

Now, we find the discriminant(D) = b2 – 4ac

D = (10)2 – 4(√3)(-8√3)

= 100 + 96

= 196

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

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

Hence, the value of x is 

x=\frac{-5+7}{\sqrt{3}}=\frac{2}{\sqrt 3}

x=\frac{-5-7}{\sqrt{3}}= \frac{-12}{\sqrt{3}} = -4\sqrt{3}

(iv) 3x2 – 2x + 2 = 0

Solution:

Given quadratic equation: 3x2 – 2x + 2 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)



On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (-2)2 – 4(3)(2)

= 4 – 24

= -20

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation does not satisfy the given condition, so it does not have real roots.

(v) 2x2 – 2√6x + 3 = 0

Solution:

Given quadratic equation: 2x2 – 2√6x + 3 = 0  ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 2, b= -2√6 and c = 3

Now, we find the discriminant(D) = b2 – 4ac

D = (-2√6)2 – 4(2)(3)

= 24 – 24

= 0

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

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

(vi) 3a2x2 + 8abx + 4b2 = 0 

Solution:

Given quadratic equation: 3a2x2 + 8abx + 4b2 = 0   ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (8ab)2 – 4(3a2)(4b2)

= 64a2b2 – 48a2b2

= 16a2b2

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get



x=\frac{-8ab \pm \sqrt{ 16a^2b^2}}{6a^2}

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

Hence, the value of x is 

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

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

(vii) 3x2 – 2√5x – 5 = 0 

Solution:

Given quadratic equation: 3x2 – 2√5x – 5 = 0    ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 3, b = 2√5 and c = -5

Now, we find the discriminant(D) = b2 – 4ac

D = (2√5 )2 – 4(3)(-5)

= 20 + 60 

= 80

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{-2\sqrt5 \pm \sqrt{80}}{2(3)}

x = \frac{-\sqrt5 \pm 2\sqrt5}{3}

Hence, the value of x is 

x = √5 /3

x = -√5 

(viii) x2 – 2x + 1 = 0

Solution:

Given quadratic equation: x2 – 2x + 1 = 0    ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  



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

Now, we find the discriminant(D) = b2 – 4ac

D = (-2)2 – 4(1)(1)

= 4 – 4

= 0 

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

x = 1

(ix) 2x2 + 5√3x + 6 = 0  

Solution:

Given quadratic equation:  2x2 + 5√3x + 6 = 0      ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 2, b = 5√3, and c = 6

Now, we find the discriminant(D) = b2 – 4ac

D = (5√3)2 – 4(2)(6)

= 75 – 48

= 27

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{-5\sqrt3 \pm \sqrt{ 27}}{2(2)} =\frac{-5\sqrt3\pm 3\sqrt 3}{4}

Hence, the value of x is 

x = -√3 /2

x = -2√3 

(x) √2x2 + 7x + 5√2 = 0    

Solution:

Given quadratic equation: √2x2 + 7x + 5√2 = 0       ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = √2, b = 7 and c = 5√2 

Now, we find the discriminant(D) = b2 – 4ac

D = (7)2 – 4(√2)(5√2)

= 49 – 40



= 9 

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{-7\pm \sqrt9}{2\sqrt2}

Hence, the value of x is 

x = -√2 

x = -5/√2 

(xi) 2x2 – 2√2x + 1 = 0   

Solution:

Given quadratic equation: 2x2 – 2√2x + 1 = 0          ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 2, b = -2√2 and c = 1

Now, we find the discriminant(D) = b2 – 4ac

D = (-2√2)2 – 4(2)(1)

= 8 – 8

= 0

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{-(-2\sqrt2)\pm \sqrt 0}{2(2)} =\frac{2\sqrt2}{4}

Hence, the value of x is 

x = 1/√2

(xii) 3x2 – 5x + 2 = 0

Solution:

Given quadratic equation: 3x2 – 5x + 2 = 0        ….(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (-5)2 – 4(3)(2)

= 25 – 24

= 1 

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{-(-5)\pm \sqrt 1}{2(3)} = \frac{5\pm1}{6}

Hence, the value of x is 

x = 1

x = 2/3

Question 3. Solve for x:

(i) \frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}, x ≠ 2, 4

Solution:

Given: \frac{x-1}{x-2}+\frac{x-3}{x-4}=3\frac{1}{3}

We can also write as



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

6x2 – 30x + 30 = 10x2 – 60x + 80

4x2 – 30x + 50 = 0

2x2 – 15x + 25 = 0 …(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (-15)2 – 4(2)(25)

= 225 – 200

= 25

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

Hence, the value of x is 

x = 5

x = 5/2

(ii) x + 1/x = 3, x ≠ 0 

Solution:

Given: x + 1/x = 3

We can also write as

x2 – 3x + 1 = 0 …(1)

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (-3)2 – 4(1)(1)

= 9 – 4

= 5 

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

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

Hence, the value of x is 

x=\frac{3+\sqrt5}{2}

x=\frac{3-\sqrt5}{2}

(iii) \frac{16}{x}-1 = \frac{15}{x+1}, x ≠ 0, -1

Solution:

Given: \frac{16}{x}-1 = \frac{15}{x+1}

We can also write as

\frac{16-x}{x}= \frac{15}{x+1}  

(16 – x)(x + 1) = 15x

15x + 16 – x2 – 15x = 0 

16 – x2 = 0

x2 – 16 = 0 ……(1) 



As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

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

Now, we find the discriminant(D) = b2 – 4ac

D = (0)2 – 4(1)(-16)

= 64

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{0\pm \sqrt {64}}{2(1)} =\frac{\pm8}{2}=\pm4

Hence, the value of x is 

x = ±4

(iv) \frac{1}{x}+\frac{2}{2x-3} = \frac{1}{x-2}, x ≠ 0, 3/2, 2

Solution:

Given: \frac{1}{x}+\frac{2}{2x-3} = \frac{1}{x-2}

We can also write as

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

(x – 2)(4x – 3) = x(2x – 3)

x2 – 4x + 3 = 0 ……(1) 

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 1, b = -4, and c = 3

Now, we find the discriminant(D) = b2 – 4ac

D = (-4)2 – 4(1)(3)

= 4

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{4\pm \sqrt {4}}{2(1)} =\frac{\pm6}{2}=\pm3

Hence, the value of x is 

x = ±3

(v) \frac{1}{x}+\frac{2}{2x-3} = \frac{1}{x-2}, x ≠ 3, -5

Solution:

Given: \frac{1}{x}+\frac{2}{2x-3} = \frac{1}{x-2}

We can also write as

\frac{(x+5)-(x-3)}{(x-3)(x+5)}= \frac{1}{6}  



(x – 3)(x + 5) = 6 x 8

x2 + 2x – 63 = 0 ……(1) 

As we know that the general form of quadratic equation is 

ax2 + bx + c = 0    ….(2)

On comparing eq(1) and (2), we get  

Here, a = 1, b = 2, and c = -63

Now, we find the discriminant(D) = b2 – 4ac

D = (2)2 – 4(1)(-63)

= 256

As we know that for a quadratic equation having real root must satisfy the D >= 0 

Here, our equation satisfies the given condition, so it has real roots.

Now we find the real roots using the given formula:

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

Put the values of b, D, a in the given formula, we get

x=\frac{2\pm \sqrt {256}}{2(1)} =\frac{\pm18}{2}=\pm9

Hence, the value of x is 

x = ±9




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