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 get

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

∴ –2i = a² – b² + 2abi

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`