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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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4  
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
1  
With $x=\frac14(\color{red}{+}7\pm\sqrt{-39})$, in fact. –  Did Aug 17 '12 at 18:34

1 Answer 1

up vote 2 down vote accepted

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

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