Chapter 14: Quadratic Equations – Exercise 14.2

Quadratic Equations – Exercise 14.2 – Q.2(iii)

(2 + i)x² - (5 – i)x + 2(1 – i) = 0

⟹ x [2 + i)x – 2] – (1 – i) [(2 + i)x – 2] = 0

⟹ [x -(1 – i) ][(2 + i) x – 2] = 0

either [x – (1 – i)] = 0 or [(2 + i) × – 2] = 0

Quadratic Equations – Exercise 14.2 – Q.2(iv)

x² - (2 + i)x - (1 - 7i) = 0

⟹ x² - (2 + i)x - (1 - 7i) = 0

⟹ x² - (3 - i)x + (1 - 2i)x - (1 - 7i) = 0

⟹ x(x - (3 - i)) + (1 - 2i)(x - (3 - i)) = 0

⟹ [x + (1 - 2i)][x - (3 - i)] = 0

⟹ x =  – 1 + 2i, 3 –  i
⟹ x =  – 1 + 2i, 3 – i

 

Quadratic Equations – Exercise 14.2 – Q.2(v)

ix² - 4x - 4i = 0

⟹ ix² + 4i²x + 4i³ = 0     [∵ i² = – 1]

⟹ x² + 4ix + 4i² = 0

⟹ x² + 2ix + 2ix + 4i² = 0

⟹ x(x + 2i) + 2i(x + 2i) = 0

⟹ (x + 2i)(x + 2i)

∴  x = – 2i, – 2i

 

Quadratic Equations – Exercise 14.2 – Q.2(vi)

x² + 4ix – 4 = 0

⟹ x² + 4ix + 4i² = 0         [∵  i² = – 1]

⟹ x² + 2ix + 2ix + 4i² = 0

⟹ x(x + 2i)+2i(x + 2i) = 0

⟹ (x + 2i)(x + 2i) = 0

⟹ x = – 2i, – 2i

 

Quadratic Equations – Exercise 14.2 – Q.2(vii)

Comparing the given equation with the general form

Substituting a and b in,

⟹ – 15 + 8i = (a + bi)²

⟹ –15 + 8i = (a²- b² + 2abi

⟹a² - b² = – 15 and 2abi = 8i

Now (a² + b²)² = (a² - b²)² + 4a²b²

⟹(a² + b²)² = (-15)² + 64 = 289

⟹ a² + b² = 17

Solving a² - b² = – 15 and a² + b² = 17, we get

a² = 1 and b² = 16

⟹ a = ± 1 and b = ± 4

⟹ a = ± 1 and b = ± 4

⟹ a = 1, b = 4 or a = – 1 b = – 4

 

Quadratic Equations – Exercise 14.2 – Q.2(viii)

x² - x + (1 + i) = 0

x² - x + (1 + i) = 0

x² - ix -(1 - i)x + i(1 - i) = 0

(x - i)(x - (1 - i)) = 0

x = i, 1 - i

 

Quadratic Equations – Exercise 14.2 – Q.2(ix)

We will apply discriminate rule on ax² + bx + c = 0,

Now,

ix² - x + 12i = 0

 

Quadratic Equations – Exercise 14.2 – Q.2(x)

We will apply discriminate rule on ax² + bx + c = 0,

Now,