# How do you complete the square to solve -2x^2 - 7x + 4 = 0?

Jun 1, 2015

f(x) = -2(x^2 + 7x/2 + 49/16) - 49/16 - 2)= 0

$- 2 {\left(x + \frac{7}{4}\right)}^{2} + \frac{98}{16} + 4$

$- 2 {\left(x + \frac{7}{4}\right)}^{2} = - \frac{98}{16} - 4 = - \frac{162}{16}$

${\left(x + \frac{7}{4}\right)}^{2} = \frac{81}{16} \to \left(x + \frac{7}{4}\right) \pm \frac{9}{4}$

$x = - \frac{7}{4} \pm \frac{9}{4}$ -> x = -4 and x = 1/2

NOTE. This method of completing the squares is unadvised when b is an odd number and a is negative. Completing the squares in this case is very challenging and easy to make errors/mistakes.
Solving by the new Transforming method (Google, Yahoo Search) is a lot simpler.
y = -(2x^2 + 7x - 4) = 0 (1)
y' = -(x^2 + 7x - 8) (2) -> a.c = -8 --> factor: (-1, 8) -> sum = b
Two real roots of (2): 1 and -8
Two real roots of (1): 1/2 and -4