[Math] Completing the Square with an A greater than 1

algebra-precalculuscompleting-the-squarefactoring

I need some assistance with a specific problem where the equation given is

$-3x^2-3x+9=0$

I have divided everything by $-3$ to get

$x^2+x-3=0$

Then I move the $3$ to the other side

$x^2+x=3$

Then I complete the square, in this case, what I need to add to complete it is $.25$

$x^2+x+.25=3.25$

Then I factor

$(x+.5)^2=3.25$

Then I take the square root

$x+.5=\sqrt{3.25}$

$\sqrt{3.25}$ is one of the answers to the problem are $x = -1/2,+/-\sqrt{13}/2$
How can the -1/2 part be found?

Any help is appreciated, thank you

Best Answer

When you take the square root, you should account for both possibilities:

$$x+.5 = \pm \sqrt{3.25}.$$

Then subtract $.5$ from both sides:

$$x = -.5 \pm \sqrt{3.25}.$$

If you convert your decimals into fractions, you have

$$x = -\frac{1}{2} \pm \sqrt{\frac{13}{4}} = -\frac{1}{2} \pm \frac{\sqrt{13}}{2}.$$