Question

Solve the following quadratic equation.

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

Answer 1

The 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 = `6 sqrt (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 - 24 sqrt (2) i`

= `18 - 12 sqrt (2) i - 4`   ...[∵ i² = – 1]

= `14 - 12 sqrt (2) i`

So, the given equation has complex roots.

These roots are given by

x = `(-b ± sqrt (b^2 - 4 ac))/(2 a)`

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

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

Let `sqrt(14 - 12 sqrt (2) i)` = a + bi, where a, b ∈ R

Squaring on both sides, we get

`14 - 12 sqrt (2) i` = a² + i²b² + 2abi

`14 - 12 sqrt (2) i` = a² – b² + 2abi

Equating real and imaginary parts, we get

a² – b² = 14 and 2ab = `-12 sqrt 2`

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

∴ `a^2 - ((-6 sqrt (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 = `± 3 sqrt (2)`

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

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

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

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

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

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

∴ x = `3 sqrt (2)` or x = 2i