0

Guest

Courses New
Book a free demo of live class

Chapter 14: Quadratic Equations – Exercise 14.2

Quadratic Equations – Exercise 14.2 – Q.1(i)

x2 + 10ix – 21 = 0

⟹ x2 + 10ix + 21i2 = 0              [∵  i2 = – 1]

⟹ x2 + 7ix + 3ix + 21i2 = 0

⟹ x(x + 7i) +3i(x + 7i) = 0

⟹ (x + 3i)(x + 7i) = 0

∴ x = – 3i, – 7i

 

Quadratic Equations – Exercise 14.2 – Q.1(ii)

x2 + (1 – 2i)x – 1i = 0

⟹ x2 + x – 2i – 2i = 0

⟹ x(x + 1) – 2i(x + 1) = 0

⟹ (x – 2i)(x + 1) = 0

⟹ x = 2i, – 1

 

Quadratic Equations – Exercise 14.2 – Q.1(iii)

Quadratic Equations – Exercise 14.2 – Q.1(iii)

 

Quadratic Equations – Exercise 14.2 – Q.1(iv)

6x2 - 17i x – 12 = 0

⟹ 6x2 - 17i x + 12i2 = 0              [∵ i2 = – 1]

⟹ 6x2 - 9ix – 8ix + 12i2 = 0

⟹3x(2x – 3i) – 4i(2x – 3i) = 0

⟹ (3x – 4i)(2x – 3i) = 0

⟹ x =4/3i or 3/2i

 

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

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

 

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

x2 – (5 – i)x + (18 + i) = 0

⟹ x2 - 5x – ix + 18 + i = 0

⟹ x2 - (3 – 4i)x – (2 + 3i)x + (18 + i) = 0

⟹ x(x – (3 – 4i)) – (2 + 3i)(x – (3 – 4i)) = 0

⟹ (x – (2 + 3i))(x – (3 – 4i)) = 0

⟹ x = 2 + 3i or 3 – 4i

 

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(iii)

 

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)

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,

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

Now,

ix2 - x + 12i = 0

 

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

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

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

Now,


TOP Your EXAMS!

Upto 50% Scholarship on Live Classes

Course Features

  • Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Book a Live class Live courses Ask question
r