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

1 Answer
Jul 13, 2017

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