Question

How do you solve `2x^2-3x-9=0`?

Answer 1

`x=-3/2" "` or `" "x=3`

Explanation:

Given:

`2x^2-3x-9=0`

Note that this quadratic is in the standard form:

`ax^2+bx+c=0`

with `a=2`, `b=-3` and `c=-9`.

Its discriminant `Delta` is given by the formula:

`Delta = b^2-4ac = (color(blue)(-3))^2-4(color(blue)(2))(color(blue)(-9)) = 9+72 = 81 = 9^2`

Since this is positive and a perfect square, we can deduce that this quadratic has rational zeros that we can find by factoring with integer coefficients...

Use an AC method:

Look for a pair of factors of `AC=2*9=18` which differ by `B=3`.

The pair `6, 3` works in that `6*3=18` and `6-3=3`.

Use this pair to split the middle term and factor by grouping:

`0 = 2x^2-3x-9`

`color(white)(0) = (2x^2-6x)+(3x-9)`

`color(white)(0) = 2x(x-3)+3(x-3)`

`color(white)(0) = (2x+3)(x-3)`

Hence zeros:

`x=-3/2" "` or `" "x=3`