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
On
$$\left(\frac a2\right)^2=\frac{a^2}{2^2}=\frac{a^2}4.$$ 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.)
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. $$ x=\pm\dfrac{\sqrt{3}}{2}a $$