Question

Solve the following quadratic equation:

x² – (5 – i)x + (18 + i) = 0

Answer 1

Given equation is x² – (5 – i)x + (18 + i) = 0

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

a = 1, b = – (5 – i), c = 18 + i

Discriminant = b² – 4ac

= [–(5 – i)]² – 4 × 1 × (18 + i)

= 25 – 10i + i² – 72 – 4i

= 25 – 10i – 1 – 72 – 4i        ...[∵ i² = – 1]

= – 48 – 14i

So, the given equation has complex roots.

These roots are given by

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

= `(-[-(5 - "i")] ± sqrt(-48 - 14"i"))/(2(1)`

= `((5 - "i") ± sqrt(-48 - 14"i"))/2`

Let `sqrt(-48 - 14"i")` = a + bi, where a, b ∈ R

Squaring on both sides, we get

– 48 – 14i = a² + b²i² + 2abi

∴ – 48 – 14i = (a² – b²) + 2abi       ...[∵ i² = – 1]

Equating real and imaginary parts, we get

a² – b² = – 48 and 2ab = – 14

∴ a² – b² = – 48 and b = `(-7)/"a"`

∴ `"a"^2 - ((-7)/"a")^2` = – 48

∴ `"a"^2 - 49/"a"^2` = – 48

∴ a⁴ –  49 = –  48a²

∴ a⁴ + 48a² – 49 = 0

∴ (a² + 49)(a² – 1) = 0

∴ a² = – 49 or a² = 1

But a ∈ R

∴ a² ≠ – 49

∴ a² = 1

∴ a = ± 1

When a = 1, b = `-7/1` = –  7

When a = –  1, b = `(-7)/(-1)` = 7

∴ `sqrt(-48 - 14"i")` = ± (1 –  7i)

∴ x = `((5 - "i") ± (1 - 7"i"))/2`

∴ x = `(5 - "i" + 1 - 7"i")/2 or x  = (5 - "i" - 1 + 7"i")/2`

∴ x = 3 – 4I or x = 2 + 3i