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

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

### (i) 2x^{2 }– 5x + 3 = 0

**Solution:**

Given quadratic equation: 2x

^{2 }– 5x + 3 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-5)

^{2}– 4(2)(3)= 25 – 24

= 1

Hence, the discriminant of given quadratic equation is 1

### (ii) x^{2 }+ 2x + 4 = 0

**Solution:**

Given quadratic equation: x

^{2 }+ 2x + 4 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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, 2x

^{2 }– 3x + 1 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-3)

^{2}– 4(2)(1)= 9 – 8

= 1

Hence, the discriminant of given quadratic equation is 1

### (iv) x^{2 }– 2x + k = 0

**Solution:**

Given quadratic equation: x

^{2 }– 2x + k = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-2)

^{2}– 4(1)(k)= 4 -4k

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

### (v) √3x^{2} + 2√2x – 2√3 = 0

**Solution:**

Given quadratic equation:√3x

^{2}+ 2√2x – 2√3 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (2√2)

^{2}– 4√3(-2√3)= 8 + 24

= 32

Hence, the discriminant of given quadratic equation is 32

### (vi) x^{2 }– x + 1 = 0

**Solution:**

Given quadratic equation: x

^{2 }– x + 1 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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) 16x^{2} = 24x + 1

**Solution:**

Given quadratic equation: 16x

^{2 }– 24x – 1 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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) x^{2 }+ x + 2 = 0

**Solution:**

Given quadratic equation: x

^{2 }+ x + 2 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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) √3x^{2} + 10x – 8√3 = 0

**Solution:**

Given quadratic equation: √3x

^{2}+ 10x – 8√3 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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:

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

Hence, the value of x is

### (iv) 3x^{2} – 2x + 2 = 0

**Solution:**

Given quadratic equation: 3x

^{2}– 2x + 2 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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) 2x^{2} – 2√6x + 3 = 0

**Solution:**

Given quadratic equation: 2x

^{2}– 2√6x + 3 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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:

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

### (vi) 3a^{2}x^{2 }+ 8abx + 4b^{2 }= 0

**Solution:**

Given quadratic equation: 3a

^{2}x^{2 }+ 8abx + 4b^{2 }= 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ bx + c = 0 ….(2)On comparing eq(1) and (2), we get

Here, a = 3a

^{2}, b = 8ab and c = 4b^{2}Now, we find the discriminant(D) = b

^{2 }– 4acD = (8ab)

^{2}– 4(3a^{2})(4b^{2})= 64a

^{2}b^{2}– 48a^{2}b^{2}= 16a

^{2}b^{2}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) 3x^{2} – 2√5x – 5 = 0

**Solution:**

Given quadratic equation: 3x

^{2}– 2√5x – 5 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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:

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

Hence, the value of x is

x = √5 /3

x = -√5

### (viii) x^{2 }– 2x + 1 = 0

**Solution:**

Given quadratic equation: x

^{2 }– 2x + 1 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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) 2x^{2 }+ 5√3x + 6 = 0

**Solution:**

Given quadratic equation: 2x

^{2 }+ 5√3x + 6 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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:

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

Hence, the value of x is

x = -√3 /2

x = -2√3

### (x) √2x^{2 }+ 7x + 5√2 = 0

**Solution:**

Given quadratic equation: √2x

^{2 }+ 7x + 5√2 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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:

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

Hence, the value of x is

x = -√2

x = -5/√2

### (xi) 2x^{2 }– 2√2x + 1 = 0

**Solution:**

Given quadratic equation: 2x

^{2 }– 2√2x + 1 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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:

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

Hence, the value of x is

x = 1/√2

### (xii) 3x^{2 }– 5x + 2 = 0

**Solution:**

Given quadratic equation: 3x

^{2 }– 5x + 2 = 0 ….(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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

### Question 3. Solve for x:

### (i) , x ≠ 2, 4

**Solution:**

Given:

We can also write as

6x

^{2 }– 30x + 30 = 10x^{2 }– 60x + 804x

^{2 }– 30x + 50 = 02x

^{2 }– 15x + 25 = 0 …(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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 ≠ 0

**Solution:**

Given: x + 1/x = 3

We can also write as

x

^{2 }– 3x + 1 = 0 …(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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 ≠ 0, -1

**Solution:**

Given:

We can also write as

(16 – x)(x + 1) = 15x

15x + 16 – x

^{2 }– 15x = 016 – x

^{2 }= 0x

^{2 }– 16 = 0 ……(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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 = ±4

### (iv) , x ≠ 0, 3/2, 2

**Solution:**

Given:

We can also write as

(x – 2)(4x – 3) = x(2x – 3)

x

^{2 }– 4x + 3 = 0 ……(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (-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 = ±3

### (v) , x ≠ 3, -5

**Solution:**

Given:

We can also write as

(x – 3)(x + 5) = 6 x 8

x

^{2 }+ 2x – 63 = 0 ……(1)As we know that the general form of quadratic equation is

ax

^{2 }+ 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) = b

^{2 }– 4acD = (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 = ±9