Question

What are the zero(s) of: `3x^2 - 5x -4 = 0`?

Answer 1

`x = 5/6+-sqrt(73)/6`

Explanation:

Given:

`3x^2-5x-4 = 0`

The difference of squares identity can be written:

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

Complete the square and use this with `A=(6x-5)` and `B=sqrt(73)` as follows:

`0 = 12(3x^2-5x-4)`

`color(white)(0) = 36x^2-60x-48`

`color(white)(0) = (6x)^2-2(6x)(5)+25-73`

`color(white)(0) = (6x-5)^2-(sqrt(73))^2`

`color(white)(0) = ((6x-5)-sqrt(73))((6x-5)+sqrt(73))`

`color(white)(0) = (6x-5-sqrt(73))(6x-5+sqrt(73))`

Hence:

`6x = 5+-sqrt(73)`

Hence:

`x = 5/6+-sqrt(73)/6`