# How do you solve 4x^2 +12x=29 by completing the square?

Jan 25, 2017

#### Explanation:

Divide both sides by 4:

${x}^{2} + 3 x = \frac{29}{4}$

Add ${a}^{2}$ to both sides:

${x}^{2} + 3 x + {a}^{2} = \frac{29}{4} + {a}^{2} \text{ [1]}$

Using the pattern ${\left(x + a\right)}^{2} = {x}^{2} + 2 a x + {a}^{2}$, set the middle term of the pattern equal to the middle term on the left of equation [1]:

$2 a x = 3 x$

Divide both sides of the equation 2x:

$a = \frac{3}{2}$

Substitute $\frac{3}{2}$ for "a" on both sides of equation [1]:

${x}^{2} + 3 x + {\left(\frac{3}{2}\right)}^{2} = \frac{29}{4} + {\left(\frac{3}{2}\right)}^{2} \text{ [2]}$

We know that the left side of equation [2] is a perfect square, therefore, we can substitute the left side of the pattern with $a = \frac{3}{2}$ into equation [2]:

${\left(x + \frac{3}{2}\right)}^{2} = \frac{29}{4} + {\left(\frac{3}{2}\right)}^{2} \text{ [3]}$

Simplify the right side of equation [3]:

${\left(x + \frac{3}{2}\right)}^{2} = \frac{38}{4} \text{ [4]}$

Use the square root on both sides:

$x + \frac{3}{2} = \pm \frac{\sqrt{38}}{2} \text{ [5]}$

Subtract $\frac{3}{2}$ form both sides:

$x = - \frac{3}{2} \pm \frac{\sqrt{38}}{2} \text{ [6]}$

Split equation [6] into two equations:

$x = - \frac{3 + \sqrt{38}}{2} \mathmr{and} x = - \frac{3 - \sqrt{38}}{2}$