Quadratic Equations and Completing the Square

[Black_Mamba]

I am stuck on this problem. I can reach the correct answer with the quadratic formula, but not with the method suggested (completing the square). Thanks in advance.

The problem is:
$$2x^2+8x+1=0$$

This is what I tried:
$$2x^2+8x=-1$$
$$2(x^2+4x)=-1$$
$$2(x^2+4x+4)=-1+8$$
$$2(x+2)^2=7$$
$$(x+2)^2=\frac{7}{2}$$
$$x=-2+\sqrt{\frac{7}{2}}$$

 

[courtrigrad]

$$2x^{2} + 8x + 1 = 0$$. First divide the equation by $$2$$. So we get: $$x^{2} + 4x + \frac{1}{2} = 0$$.

So $$x^{2} + 4x = -\frac{1}{2}$$.

$$x^{2} + 4x + 4 = \frac{7}{2}$$
$$(x+2)^{2} = \frac{7}{2}$$.
$$x+2 = \sqrt{\frac{7}{2}}$$.
$$x = \sqrt{\frac{7}{2}} -2$$.

 

[Black_Mamba]

Thanks for the quick reply, but the answer is supposed to end up being

$$x=-2+{\frac{\sqrt14}{2}}$$

 

[courtrigrad]

$$\sqrt{\frac{7}{2}} = \frac{\sqrt{7}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{14}}{2}$$. It is the same answer. They just rationalized the denominator.

 

[Black_Mamba]

Ohh.. I see. Silly me. Thank you!