Question -
Answer -
(i) x2 +10ix – 21 = 0
Given: x2 +10ix – 21 = 0
x2 +10ix – 21 × 1 = 0
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
x2 +10ix – 21(–i2) = 0
x2 +10ix + 21i2 = 0
x2 +3ix + 7ix + 21i2 = 0
x(x + 3i) + 7i(x + 3i)= 0
(x + 3i) (x + 7i) = 0
x + 3i = 0 or x + 7i =0
x = –3i or –7i
∴ The roots of thegiven equation are –3i, –7i
(ii) x2 +(1 – 2i)x – 2i = 0
Given: x2 +(1 – 2i)x – 2i = 0
x2 + x– 2ix – 2i = 0
x(x + 1) – 2i(x + 1) =0
(x + 1) (x – 2i) = 0
x + 1 = 0 or x – 2i =0
x = –1 or 2i
∴ The roots of thegiven equation are –1, 2i
(iii) x2 –(2√3 + 3i) x + 6√3i = 0
Given: x2 –(2√3 + 3i) x + 6√3i = 0
x2 –(2√3x + 3ix) + 6√3i = 0
x2 –2√3x – 3ix + 6√3i = 0
x(x – 2√3) – 3i(x –2√3) = 0
(x – 2√3) (x – 3i) = 0
(x – 2√3) = 0 or (x –3i) = 0
x = 2√3 or x = 3i
∴ The roots of thegiven equation are 2√3, 3i
(iv) 6x2 –17ix – 12 = 0
Given: 6x2 –17ix – 12 = 0
6x2 –17ix – 12 × 1 = 0
We know, i2 =–1 ⇒ 1 = –i2
By substituting 1 = –i2 inthe above equation, we get
6x2 –17ix – 12(–i2) = 0
6x2 –17ix + 12i2 = 0
6x2 –9ix – 8ix + 12i2 = 0
3x(2x – 3i) – 4i(2x –3i) = 0
(2x – 3i) (3x – 4i) =0
2x – 3i = 0 or 3x – 4i= 0
2x = 3i or 3x = 4i
x = 3i/2 or x = 4i/3
∴ The roots of thegiven equation are 3i/2, 4i/3