# How do you simplify 5sqrt(16x^3) - 2xsqrt(9x)?

May 26, 2015

First note that $x \ge 0$ - otherwise the square roots will not have real values.

That having been said:

$5 \sqrt{16 {x}^{3}} - 2 x \sqrt{9 x}$

$= 5 \sqrt{{\left(4 x\right)}^{2} \cdot x} - 2 x \sqrt{{3}^{2} \cdot x}$

$= 5 \sqrt{{\left(4 x\right)}^{2}} \sqrt{x} - 2 x \sqrt{{3}^{2}} \sqrt{x}$

$= 5 \cdot 4 x \sqrt{x} - 2 x \cdot 3 \sqrt{x}$

$= 20 x \sqrt{x} - 6 x \sqrt{x}$

$= 14 x \sqrt{x}$

This derivation mostly uses $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$

Notice that I was able to say that $\sqrt{{\left(4 x\right)}^{2}} = 4 x$ because I had already established that $x \ge 0$.

If I had not established that $x \ge 0$ then I would only have been able to say $\sqrt{{\left(4 x\right)}^{2}} = 4 \left\mid x \right\mid$