Question

Solve the following quadratic equation:

x² – (2 + i) x – (1 – 7i) = 0

Answer 1

Given equation is x² – (2 + i) x – (1 – 7i) = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = – (2 + i), c = –(1 – 7i)
Discriminant = b² – 4ac
= [– (2 + i)]² – 4 x 1 – (1 – 7i)
= 4 + 4i + i² + 4 – 28i
= 4 + 4i – 1 + 4 – 28i          ...[∵ i² = – 1]
= 7 – 24i
So, the given equation has complex roots.
These roots are given by

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

= `(-[-(2 + "i")] ± sqrt(7 - 24"i"))/(2(1)`

= `((2 + "i") ± sqrt(7 - 24"i"))/2`

Let `sqrt(7 - 24"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
7 – 24i = a² + i²b² + 2abi
∴ 7 – 24i = (a² – b²) + 2abi        ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 7 and 2ab = – 24

∴ a² – b² = 7 and b = `(-12)/"a"`

∴ `"a"^2 - ((-12)/"a")^2` = 7

∴ `"a"^2 - 144/"a"^2` = 7

∴ a⁴ – 144 = 7a²
∴ a⁴ – 7a² – 144 = 0
∴ (a² – 16) (a² + 9) = 0
∴ a² = 16 or a² = – 9
But a ∈ R
∴ a² ≠ – 9
∴ a² = 16
∴ a = ± 4
When a = 4, b  = `(-12)/4` = – 3

When a = – 4, b = `(-12)/(-4)` = 3

∴ `sqrt(7 - 24"i")` = ± (4 – 3i)

∴ x = `((2 + "i") ± (4 - 3"i"))/2`

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

∴ x = 3 – i or x = – 1 + 2i.