How do you solve 2x^2-3x-9=0?
1 Answer
Jul 13, 2017
Explanation:
Given:
2x^2-3x-9=0
Note that this quadratic is in the standard form:
ax^2+bx+c=0
with
Its discriminant
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
The pair
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