**Solve the following quadratic equation: **

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

#### Solution

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

Comparing with ax^{2} + bx + c = 0, we get

a = 1, b = `-(3 sqrt(2) + 2"i"), "c" = 6sqrt(2)"i"`

Discriminant = b^{2} – 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` ...[∵ i^{2} = – 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"` = a^{2} + i^{2}b^{2} + 2abi

∴ `14 - 12sqrt(2) "i"` = (a^{2} – b^{2}) + 2abi ...[∵ i^{2} = – 1]

Equating real and imaginary parts, we get

a^{2} – b^{2} = 14 and 2ab = `-12sqrt(2)`

∴ a^{2} – b^{2} = 14 and b = `(-6sqrt(2))/"a"`

∴ `"a"^2 - ((-6sqrt(2))/"a")^2` = 14

∴ `"a"^2 - 72/"a"^2` = 14

∴ a^{4} – 72 = 14a^{2}

∴ a^{4} – 14a^{2 }– 72 = 0

∴ (a^{2} – 18) (a^{2} + 4) = 0

∴ a^{2} = 18 or a^{2} = – 4

But a ∈ R

∴ a^{2} ≠ – 4

∴ a^{2} = 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.