## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 14 (Mis Ex)

Question 14:

Find the real numbers x and y if (x – iy) (3 + 5i) is the conjugate of −6 – 24i.

Let z = (x – iy) (3 + 5i)

z = 3x + 5xi – 3yi – 5yi2 = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y) ∴ (3x + 5y) – i(5x – 3y) = −6 – 24i

Equating real and imaginary parts, we obtain

3x + 5y = −6                     ….(i)

5x – 3y = 24                      ….(ii)

Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain 9x + 15y = −18 Putting the value of x in equation (i), we obtain

3(3) + 5y = −6

⇒ 5y = −6 – 9 = −15

⇒ y = −3

Thus, the values of x and y are 3 and −3 respectively.

## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 13 (Mis Ex)

Question 13:  Let z = r cos θ + ir sin θ

i.e., r cos θ = −1/2 and r sin θ = 1/2

On squaring and adding, we obtain Therefore, the modulus and argument of the given complex number are 1/√2 and 3π/4 respectively.

## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 12 (Mis Ex)

Question 12:

Let z1 = 2 – i, z2 = −2 + i. Find z1 = 2 – i, z2 = −2 + i

(i) z1z2 = (2 – i) (−2 + i) = −4 + 2i + 2i – i2 = −4 + 4i – (−1) = −3 + 4i   ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 11 (Mis Ex)

Question 11:    ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 10 (Mis Ex)

Question 10:   ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 9 (Mis Ex)

Question 9 :

Solve the equation 21x2 – 28x + 10 = 0

The given quadratic equation is 21x2 – 28x + 10 = 0

On comparing the given equation with ax2 + bx + c = 0, we obtain

a = 21, b = −28, and c = 10

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (−28)2 – 4 × 21 × 10 = 784 – 840 = −56

Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 8 (Mis Ex)

Question 8:

Solve the equation 27x2 – 10x + 1 = 0

The given quadratic equation is 27x2 – 10x + 1 = 0

On comparing the given equation with ax2 + bx + c = 0, we obtain

a = 27, b = −10, and c = 1

Therefore, the discriminant of the given equation is

D = b2 – 4ac =(−10)2 – 4 × 27 × 1 = 100 – 108 = −8

Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 7 (Mis Ex)

Question 7:  This equation can also be written as 2x3 – 4x + 3 = 0

On comparing this equation with ax2 + bx + c = 0, we obtain

a = 2, b = −4, and c = 3

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (−4)2 – 4 × 2 × 3 = 16 – 24 = −8

Therefore, the required solutions are ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 6 (Mis Ex)

Question 6:  This equation can also be written as 9x2 – 12x + 20 = 0

On comparing this equation with ax2 + bx + c = 0, we obtain

a = 9, b = −12, and c = 20

Therefore, the discriminant of the given equation is

D = b2 – 4ac = (−12)2 – 4 × 9 × 20 = 144 – 720 = −576 ## NCERT Solution Class XI Mathematics Complex Numbers and Quadratic Equations Question 5 (Mis Ex)

Question 5:

Convert the following in the polar form :  Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ) = 1 + 1

⇒ r2(cos2 θ + sin2 θ) = 1 + 1

⇒ r2(cos2 θ + sin2 θ) = 2

⇒ r2 = 2                      [cos2 θ + sin2 θ = 1]

⇒ r = √2                  [Conventionally, r > 0]

∴ √2 cos θ = −1 and √2 sin θ = 1  Let r cos θ = −1 and r sin θ = 1

On squaring and adding, we obtain

r2(cos2 θ + sin2 θ)=1 + 1

⇒ r2(cos2 θ + sin2 θ) = 2

⇒ r2 = 2                     [cos2 θ + sin2 θ = 1]

⇒ r = √2                  [Conventionally, r > 0]

∴ √2 cos θ =−1 and √2 sin θ = 1 This is the required polar form.