Question

Solve the following quadratic equation:

(2 + i) x² – (5 – i) x + 2(1 – i) = 0

Answer 1

Given equation is

(2 + i) x² – (5 – i) x + 2(1 – i) = 0

Comparing with ax² + bx + c = 0, we geta

a = 2 + i, b = – (5 –  i), c = 2(1 - i)

Discriminant = b² –  4ac

= [–  5 –  i)]² –  4 x (2 + i) x 2(1 –  i)

= 25 –  10i + i² –  8(2 + i) (1 - i)

= 25 –  10i + i² –  8(2 –  2i + i –  i²)

= 25 –  10i –  1 - 8(2 - i + 1)          ...[∵ i² = – 1]

= 25 – 10i – 1 – 16 + 8i – 8

= – 2i

So, the given equation has complex roots.
These roots are given by

x = `(-"b" ± sqrt("b"^2 - 4"ac"))/(2"a")`

= `(-[-(5 - "i")] ± sqrt(-2"i"))/(2(2 + "i")`

= `((5 - "i") ± sqrt(-2"i"))/(2(2 + "i")`

Let `sqrt(-2"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
– 2i = a² + b²i² + 2abi
∴ 0 – 2i = (a² – b²) + 2abi      ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 0 and 2ab = – 2

∴ a² – b² = 0 and b = `-1/"a"`

∴ `"a"^2 - ((-1)/"a")^2` = 0

∴ `"a"^2 - 1/"a"^2` = 0

∴ a⁴ – 1 = 0
∴ (a² – 1)(a² + 1) = 0
∴ a² = 1 or a² = – 1
But a ∈ R
∴ a² ≠ – 1
∴ a² = 1
∴ a = ± 1
When a = 1, b = – 1
When a = – 1, b = 1
∴ `sqrt(-2"i")` = ±  (1 – i)

∴ x = `((5 - "i") ± (1 - "i"))/(2(2 + "i")`

∴ x = `(5 - "i" + 1 - "i")/(2(2 + "i")) or x = (5 - "i" - 1 + "i")/(2(2 + "i")`

∴ x =  `(6 - 2"i")/(2(2 + "i")) or x = 4/(2(2 + "i")`

∴ x =  `(2(3 - "i"))/(2(2 + "i")) or x = 2/(2 + "i")`

∴ x =  `(3 - "i")/(2 + "i") or  x = (2(2 - "i"))/((2 + "i")(2 - "i")`

∴ x = `((3 - "i")(2 - "i"))/((2 + "i")(2 - "i")) or x = (2(2 - "i"))/(4 - "i"^2)`

∴ x =  `(6 - 5"i" + "i"^2)/(4 - "i"^2) or x = (4 - 2"i")/(4 - "i"^2)`

∴ x = `(5 - 5"i")/5 or x = (4 - 2"i")/5`    ...[∵ i² = – 1]

∴ x = 1 – i or x = `4/5 - (2"i")/5`.