# Getting square root of negative in completing the square problem

I try to solve the equation $f(x) = 7x - 11 - 2x^2 = 0$ for $x$, but run into troubles. I've gone through it over and over again as well as similar problems, but can't find what I'm doing wrong.

$$f(x) = 7x - 11 - 2x^2 = 0$$ $$\iff x^2 - \frac{7}{2}x + \frac{11}{2} = 0$$ $$\iff \left(x + \frac{7}{4}\right)^2 = \left(\frac{7}{4}\right)^2 - \frac{11}{2}$$ $$\iff x + \frac{7}{4} = \pm \sqrt{\left(\frac{7}{4}\right)^2 - \frac{11}{2}}$$ $$\iff x = -\frac{7}{4} \pm \sqrt{\frac{49}{16} - \frac{88}{16}}$$ $$\iff x = -\frac{7}{4} \pm \sqrt{\frac{-39}{16}}$$

I should be able to continue but I'm stuck (seeing as it's a negative number). What am I doing wrong?

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Nothing. This equation has no real solutions. –  tomasz Aug 17 '12 at 18:28
As @tomasz said, it’s fine: the equation has no real solution. About the only useful thing you can do from here is to rewrite the solution as $x=\frac14\left(-7\pm\sqrt{-39}\right)$. –  Brian M. Scott Aug 17 '12 at 18:31
Draw the graph $y=f(x)$ - you can write a solution, but it doesn't seem to exist. Think "what if if did, what would follow from that?". Work it out for yourself as far as possible. –  Mark Bennet Aug 17 '12 at 18:34
With $x=\frac14(\color{red}{+}7\pm\sqrt{-39})$, in fact. –  Did Aug 17 '12 at 18:34

You did everything fine but your quadratic equation has no real solutions, which you could have found out way more easily had you first calculated the equation's discriminant:

$$\Delta:=b^2-4ac=7^2-4(-2)(-11)=49-88=-39<0$$

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He pretty much did just that. :) –  tomasz Aug 17 '12 at 18:33
Not the way I wrote, which is much shorter than his. Of course, as he's using the quadratic roots formula, which is only $$x_{1,2}=\frac{-b\pm\sqrt \Delta}{2a}$$ then the disciminant appears there. My point is you don't need to do all that to use the discriminant. :) –  DonAntonio Aug 17 '12 at 18:37