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!