Question

How do you solve using the completing the square method `-3x^2-5x+5 = 33`?

Answer 1

`x = -5/6+-sqrt(311)/6i`

Explanation:

Given:

`-3x^2-5x+5=33`

Adding `3x^2+5x-5` to both sides and transposing, we get:

`3x^2+5x+28 = 0`

This is in standard form:

`ax^2+bx+c = 0`

with `a=3`, `b=5` and `c=28`.

This has discriminant `Delta` given by the formula:

`Delta = b^2-4ac = 5^2-4(3)(28) = 25-336 = -311`

Since `Delta < 0`, this quadratic has no Real solutions.

We can find Complex solutions by completing the square, then using the difference of squares identity:

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

with `A=(6x+5)` and `B=sqrt(311)` as follows:

`0 = 12(3x^2+5x+28)`

`color(white)(0) = 36x^2+60x+336`

`color(white)(0) = (6x)^2+2(6x)(5)+5^2+311`

`color(white)(0) = (6x+5)^2+(sqrt(311))^2`

`color(white)(0) = (6x+5)^2-(sqrt(311)i)^2`

`color(white)(0) = ((6x+5)-sqrt(311)i)((6x+5)+sqrt(311)i)`

`color(white)(0) = (6x+5-sqrt(311)i)(6x+5+sqrt(311)i)`

Hence:

`x = 1/6(-5+-sqrt(311)i) = -5/6+-sqrt(311)/6i`