Question

Solve the following quadratic equation.

x² − (2 + i)x − (1 − 7i) = 0

Answer 1

Comparing the equation x² − (2 + i)x − (1 − 7i) = 0 

with ax² + bx + c = 0, we have,

a = 1, b = −(2 + i), c = −(1 − 7i)

∴ b² − 4ac = [−(2 +i)]² − 4(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))`

Now, we have to find `sqrt(7 - 24"i")`

Let p + qi = `sqrt(7 - 24"i")`, where p, q ∈ R

∴ (p + qi)² = 7 − 24i

∴ p² + q²i² + 2pqi = 7 − 24i

∴ (p² − q²) + 2pqi = 7 − 24i  ...[∵ i² = −1]

Equating the real and imaginary parts separately, we get,

p² − q² = 7 and 2pq = − 24

∴ q = `(-12)/"p"`

∴ `"p"^2 - ((-12)/"p")^2` = 7

∴ `"p"^2 - 144/"p"^2` = 7

∴ p⁴ − 144 = 7p²

∴ p⁴ − 7p² − 144=0

∴(p² − 16)(p² + 9) = 0

∴ p² = 16 or p² = − 9

Now, p is a real number

∴ p² ≠ − 9

∴ p² = 16

∴ p = ± 4

∴ When p = 4, q = `-12/4` = − 3

When p = − 4, q = `(-12)/(-4)` = 3

∴ `sqrt(7 - 24"i")` = ± (4 − 3i)

∴ the roots are given by

x = `((2 + "i") ± ( 4 - 3"i"))/2`

Hence, the roots of the equation are

`((2 + "i") + ( 4 - 3"i"))/2` and `((2 + "i") -  (4 - 3"i"))/2`

i.e., `(6 - 2"i")/2` and `(-2 + 4"i")/2`

i.e., 3 − i and −1 + 2i