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# Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.5

• Last Updated : 30 Apr, 2021

### (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) âˆš3x2 + 2âˆš2x – 2âˆš3 = 0

Solution:

Given quadratic equation:âˆš3x2 + 2âˆš2x – 2âˆš3 = 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 =âˆš3, b = 2âˆš2, and c = -2 – 2âˆš3

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

D = (2âˆš2)2 – 4âˆš3(-2âˆš3)

= 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

### (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:

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

Hence, the value of x is

### (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) âˆš3x2 + 10x – 8âˆš3 = 0

Solution:

Given quadratic equation: âˆš3x2 + 10x – 8âˆš3 = 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 = âˆš3, b = 10 and c = -8âˆš3

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

D = (10)2 – 4(âˆš3)(-8âˆš3)

= 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:

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

Hence, the value of x is

### (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 – 2âˆš6x + 3 = 0

Solution:

Given quadratic equation: 2x2 – 2âˆš6x + 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= -2âˆš6 and c = 3

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

D = (-2âˆš6)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:

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

### (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:

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

Hence, the value of x is

### (vii) 3x2 – 2âˆš5x – 5 = 0

Solution:

Given quadratic equation: 3x2 – 2âˆš5x – 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 = 2âˆš5 and c = -5

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

D = (2âˆš5 )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:

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

Hence, the value of x is

x = âˆš5 /3

x = -âˆš5

### (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:

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

x = 1

### (ix) 2x2 + 5âˆš3x + 6 = 0

Solution:

Given quadratic equation:  2x2 + 5âˆš3x + 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 = 5âˆš3, and c = 6

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

D = (5âˆš3)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:

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

Hence, the value of x is

x = -âˆš3 /2

x = -2âˆš3

### (x) âˆš2x2 + 7x + 5âˆš2 = 0

Solution:

Given quadratic equation: âˆš2x2 + 7x + 5âˆš2 = 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 = âˆš2, b = 7 and c = 5âˆš2

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

D = (7)2 – 4(âˆš2)(5âˆš2)

= 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:

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

Hence, the value of x is

x = -âˆš2

x = -5/âˆš2

### (xi) 2x2 – 2âˆš2x + 1 = 0

Solution:

Given quadratic equation: 2x2 – 2âˆš2x + 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 = -2âˆš2 and c = 1

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

D = (-2âˆš2)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:

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

Hence, the value of x is

x = 1/âˆš2

### (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:

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

Hence, the value of x is

x = 1

x = 2/3

### (i) , x â‰  2, 4

Solution:

Given:

We can also write as

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:

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

Hence, the value of x is

x = 5

x = 5/2

### (ii) x + 1/x = 3, x â‰  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:

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

Hence, the value of x is

### (iii) , x â‰  0, -1

Solution:

Given:

We can also write as

(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:

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

Hence, the value of x is

x = Â±4

### (iv) , x â‰  0, 3/2, 2

Solution:

Given:

We can also write as

(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:

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

Hence, the value of x is

x = Â±3

### (v) , x â‰  3, -5

Solution:

Given:

We can also write as

(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:

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

Hence, the value of x is

x = Â±9

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