# Basic calculation with root roots and power

Sorry for the boring question but I just need someone to remind me the way to calculate this:

$\displaystyle \left(\frac{a}{2}\right)^2 + x^2 = a^2$ (i used carrot sign cause i dont know how to do factorial in my mac keyboard)

The answer is $\displaystyle \frac{\sqrt {3}}{2}a$ but i forgot how to calculate this type of exercise, can someone be kind enough and help?

tnx

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I rewrote your question using $\LaTeX$, please check thatI didn't alter the meaning of the question. –  Git Gud Apr 9 '13 at 13:52
@GitGud thanks buddy, but the answer is x = to √3 divided by 2...is that the same? –  user804968 Apr 9 '13 at 13:58
@GitGud it's (a/2)² + x ² = a ² (i don't really know all the shortcuts to write it properly so i used the parenthesis) –  user804968 Apr 9 '13 at 14:02
@GitGud yes, just the answer is (√3/2).. –  user804968 Apr 9 '13 at 14:05
I hope everything is OK now, please check. –  Git Gud Apr 9 '13 at 14:06

Now I see: $\left(\dfrac{a}{2}\right)=\dfrac{a^2}{2^2}=\dfrac{a^2}{4}$, if you replace this in your original equation you get: $$\dfrac{a^2}{4}+x^2=a^2$$ If you multiply by $4$ both sides of this equation you obtain the following: $a^2+4x^2=4a^2$, and if you subtract $a^2$ from both sides you get $4x^2=3a^2$, and finally after you divide by $4$ both sides of the equation you obtain: $$x^2=\dfrac{3a^2}{4}$$ Last step is to take the square root of both sides and don't forget that squaring a quantity always produces a non-negative number independently of the sign of original quantity, so we will get two answers that will only differ by the sign. $$x=\pm\dfrac{\sqrt{3}}{2}a$$
$$\left(\frac a2\right)^2=\frac{a^2}{2^2}=\frac{a^2}4.$$ Thus, using the difference of squares identity $$y^2-z^2=(y-z)(y+z),$$ the following are equivalent: $$\left(\frac a2\right)^2+x^2=a^2\\x^2+\frac{a^2}4-a^2=0\\x^2-\frac{3a^2}4=0\\x^2-\left(\frac{a\sqrt{3}}2\right)^2=0\\\left(x-\frac{a\sqrt{3}}2\right)\left(x+\frac{a\sqrt{3}}2\right)=0.$$ Now what can you conclude from that? (Incidentally, the answer you mention is incorrect.)