Question

How do you solve `7x^2 - 12x + 16 = 0`?

Answer 1

`x = 6/7+-2/7sqrt(19)i`

Explanation:

The given quadratic equation:

`7x^2-12x+16 = 0`

is in the form:

`ax^2+bx+c = 0`

with `a=7`, `b=-12` and `c=16`

This has discriminant `Delta` given by the formula:

`Delta = b^2 - 4ac = (-12)^2-4(7)(16) = 144-448 = -304`

Since `Delta < 0`, this quadratic equation has no Real roots, only Complex one.

We can still find the roots by completing the square.

The difference of squares identity can be written:

`A^2-B^2=(A-B)(A+B)`

We use this with `A=(7x-6)` and `B=2sqrt(19)i` as follows:

`0 = 7(7x^2-12x+16)`

`color(white)(0) = 49x^2-84x+112`

`color(white)(0) = 49x^2-84x+36+76`

`color(white)(0) = (7x)^2-2(7x)(6)+(6)^2+(2sqrt(19))^2`

`color(white)(0) = (7x-6)^2-(2sqrt(19)i)^2`

`color(white)(0) = ((7x-6)-2sqrt(19)i)((7x-6)+2sqrt(19)i)`

`color(white)(0) = (7x-6-2sqrt(19)i)(7x-6+2sqrt(19)i)`

Hence:

`x = 1/7(6+-2sqrt(19)i) = 6/7+-2/7sqrt(19)i`