# The equations 2x^2 +3x=4 is rewritten in the form 2(x-h)^2+ q=0. What is the value of q?

Jan 7, 2017

$q = - \frac{41}{8}$

#### Explanation:

You would get the equivalent:

1) by subtracting 4:

$2 {x}^{2} + 3 x - 4 = 0$

2) by factorizing 2:

$2 \left({x}^{2} + \frac{3}{2} x - 2\right) = 0$

3) since

${x}^{2} + \frac{3}{2} x - 2 = {x}^{2} + \frac{3}{2} x \textcolor{red}{+ \frac{9}{16} - \frac{9}{16}} - 2$

and the first three terms are the squared binomial

${\left(x + \frac{3}{4}\right)}^{2}$,

you get:

$2 \left({\left(x + \frac{3}{4}\right)}^{2} - \frac{9}{16} - 2\right) = 0$

and then

$2 {\left(x + \frac{3}{4}\right)}^{2} + 2 \left(- \frac{9}{16} - 2\right) = 0$

where

$q = - \frac{9}{8} - 4 = - \frac{41}{8}$