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

1 Answer
George C.
May 22, 2018

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

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