Question

Solve the following quadratic equation:

`x^2 - (3 sqrt(2) + 2"i") x + 6 sqrt(2)"i"` = 0

Answer 1

Given equation is `x^2 - (3 sqrt(2) + 2"i") x + 6 sqrt(2)"i"` = 0
Comparing with ax² + bx + c = 0, we get
a = 1, b = `-(3 sqrt(2) + 2"i"), "c" = 6sqrt(2)"i"`
Discriminant = b² – 4ac
= `[-(3 sqrt(2) + 2"i")]^2 - 4 xx 1 xx 6 sqrt(2) "i"`

= `18 + 12 sqrt(2) "i" + 4"i"^2 - 24sqrt(2)"i"`

= `18 - 12sqrt(2) "i" - 4`       ...[∵ i² = – 1]
= `14 - 12sqrt(2)"i"`
So, the given equation has complex roots.
These roots are given by

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

= `(-[-(3sqrt(2) + 2"i")] ± sqrt(14 - 12sqrt(2)"i"))/(2(1)`

= `((3sqrt(2) + 2"i") ± sqrt(14 - 12sqrt(2)"i"))/2`

Let `sqrt(14 - 12sqrt(2)"i")` = a + bi, where a, b ∈ R
Squaring on both sides, we get
`14 - 12sqrt(2) "i"` = a² + i²b² + 2abi
∴ `14 - 12sqrt(2) "i"` = (a² – b²) + 2abi    ...[∵ i² = – 1]
Equating real and imaginary parts, we get
a² – b² = 14 and 2ab = `-12sqrt(2)`

∴ a² – b² = 14 and b = `(-6sqrt(2))/"a"`

∴ `"a"^2 - ((-6sqrt(2))/"a")^2` = 14

∴ `"a"^2 - 72/"a"^2` = 14

∴ a⁴ – 72 = 14a²
∴ a⁴ –  14a² –  72 = 0
∴ (a² –  18) (a² + 4) = 0
∴ a² = 18 or a² = – 4
But a ∈ R
∴ a² ≠ – 4
∴ a² = 18
∴ a = ± `3sqrt(2)`
When a = `3sqrt(2), "b" = (-6sqrt(2))/(3sqrt(2)` = – 2

When a = `-3sqrt(2), "b" = (-6sqrt(2))/(-3sqrt(2)` = 2

∴ `sqrt(14 - 12sqrt(2)"i") = ± (3sqrt(2) - 2i")`

∴ x = `((3sqrt(2) + 2"i") ± (3sqrt(2) - 2"i"))/2`

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

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

∴ x = `3sqrt(2)` or x = 2i.