Question

Solve the following quadratic equation.

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

Answer 1

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

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

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

Discriminant = b² – 4ac

= `[-(3sqrt(2) + 2"i")]^2 - 4 xx 1 xx 6sqrt(2)"i"`

= `18 + 12sqrt(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

Equating real and imaginary parts, we get

a² – b² = 14 and 2ab = `-12sqrt2`

∴ 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) - 2"i")`

∴ 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