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How do I solve the following equation?

$$x^2 + 10 = 15$$

Here's how I think this should be solved. \begin{align*} x^2 + 10 - 10 & = 15 - 10 \\ x^2 & = 15 - 10 \\ x^2 & = 5 \\ x & = \sqrt{5} \end{align*} I was thinking that the square root of 5 is iregular repeating 2.23606797749979 number. 2.236 multipled by itself equals 5ish.

I've also seen another equation like this: \begin{align*} x^2 & = 4 \\ x^2 + 4 & = 0 \\ (x - 2)(x + 2) & = 0 \\ x & = 2 \text{ or } -2 \end{align*} So I guess I could near the end of my equation do the following:

$$x^2 + 5 = 0$$

and then go from there?

Is my first attempt at solving correct?

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Going from $x^2=4$ to $x^2+4=0$ is wrong. From $x^2=4$ you can go to $x^2-4=0$. –  André Nicolas Nov 28 '12 at 0:23
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2 Answers

up vote 3 down vote accepted

Your equation should be corrected at the second line to be $$x^2 -4 = 0$$ and similarly $$x^2-5=0$$. Then $$(x-\sqrt{5})(x+\sqrt{5})=0$$ which implies $$x=\sqrt{5}$$ or $$x=-\sqrt{5}$$ The answer being irrational doesn't matter, as $x$ is in the real number set (or can even be complex number set, a quadratic equation always have a solution in the complex number set).

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Awesome thanks. I had a feeling that I was partly correct. –  developerdoug Nov 28 '12 at 15:33
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The idea is there, however be aware of the fact that $x^2 = a$ is equivalent to $x = \sqrt{a}$ OR $x = -\sqrt{a}$.

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