Question

Solve the following quadratic equation.

x² − (5 − i) x + (18 + i) = 0

Answer 1

Comparing the equation x² − (5 − i) x + (18 + i) = 0

with ax² + bx + c = 0, we have,

a = 1, b = −(5 − i), c = 18 + i

∴ 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"a")`

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

= `((5 - "i") ± sqrt(-48 - 14"i"))/2`

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

Squaring on both sides, we get

−48 − 14i = a² + b²i² + 2abi

∴ −48 − 14i = a² − b² + 2abi

Equating the real and imaginary parts separately, 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

Now, a is a real number

∴ 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)

∴ the roots are given by

x = `((5 - "i") ± ( 1 - 7"i"))/2`

Hence, the roots of the equation are

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

i.e., `x = (5 - "i" + 1 - 7"i")/2` and `x = (5 - "i" - 1 + 7"i")/2`

i.e., x = 3 − 4i and x = 2 + 3i