(50x+25/x^2)/((2)sqrt[25x^2-(25/x)]
simplified = 5(2x^3+1)/((2)sqrt[x^3(x^3-1))]
I need help on simplifying top to bottom - cannot figure how to get there!!!
Hello
First of all, thank you very much because you put the brackets just like you had to do !
Now, the problem ^^
$\displaystyle F=\frac{50x+\frac{25}{x^2}}{2 \sqrt{25x^2-\frac{25}{x}}}$
You can notice that in the solution, there is no more 25 or 50, etc... So we'll simplify by factoring as much as possible
$\displaystyle F=\frac{25 \left(2x+\frac{1}{x^2}\right)}{2 \cdot \sqrt{25 \left(x^2-\frac 1x\right)}}$
But $\displaystyle \sqrt{25}=5$ so :
$\displaystyle F=\frac{5 \cdot 5 \cdot \left(2x+\frac{1}{x^2}\right)}{2 \cdot 5 \cdot \sqrt{x^2-\frac 1x}}$
Simplify by 5 :
$\displaystyle F=\frac{5 \cdot \left({\color{red}2x+\frac{1}{x^2}}\right)}{2 \sqrt{x^2-\frac 1x}}$
Now, you can notice that the red term is very similar to what you want to get : $\displaystyle 2x^3+1$.
Actually, $\displaystyle 2x^3+1=x^2 \cdot \left({\color{red}2x+\frac{1}{x^2}}\right)$.
This gives you the trick : multiply the denominator & the numerator by $\displaystyle x^2$
---> $\displaystyle F=\frac{5 \cdot (2x^3+1)}{2 \cdot x^2 \cdot \sqrt{x^2-\frac 1x}}$
But $\displaystyle x^2=\sqrt{x^4}$.
$\displaystyle \implies F=\frac{5 \cdot (2x^3+1)}{2 \cdot \sqrt{x^4 \left(x^2-\frac 1x\right)}}$
$\displaystyle F=\frac{5 \cdot (2x^3+1)}{2 \cdot \sqrt{x^3(x^3-1)}}$
Is it clear enough ?