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Quadratic Equations and Completing the Square

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:
[tex]2x^2+8x+1=0[/tex]

This is what I tried:
[tex]2x^2+8x=-1[/tex]
[tex]2(x^2+4x)=-1[/tex]
[tex]2(x^2+4x+4)=-1+8[/tex]
[tex]2(x+2)^2=7[/tex]
[tex](x+2)^2=\frac{7}{2}[/tex]
[tex]x=-2+\sqrt{\frac{7}{2}}[/tex]
 
Last edited:
1,233
1
[tex] 2x^{2} + 8x + 1 = 0 [/tex]. First divide the equation by [tex] 2 [/tex]. So we get: [tex] x^{2} + 4x + \frac{1}{2} = 0 [/tex].

So [tex] x^{2} + 4x = -\frac{1}{2} [/tex].

[tex] x^{2} + 4x + 4 = \frac{7}{2} [/tex]
[tex] (x+2)^{2} = \frac{7}{2} [/tex].
[tex] x+2 = \sqrt{\frac{7}{2}} [/tex].
[tex] x = \sqrt{\frac{7}{2}} -2 [/tex].
 
Thanks for the quick reply, but the answer is supposed to end up being

[tex]x=-2+{\frac{\sqrt14}{2}}[/tex]
 
1,233
1
[tex] \sqrt{\frac{7}{2}} = \frac{\sqrt{7}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{14}}{2} [/tex]. It is the same answer. They just rationalized the denominator.
 
Last edited:
Ohh.. I see. Silly me. Thank you!
 

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