# How do you simplify sqrt(9x^4)?

Apr 29, 2016

$3 {x}^{2}$

#### Explanation:

$9 = {3}^{2} \text{ and } {x}^{4} = {\left({x}^{2}\right)}^{2}$

Multiply any number by 1 and you do not change that number's value. I am including the value 1 just to demonstrate a process.

So write as:" sqrt(1xx3^2xx(x^2)^2)

Taking all squared values outside the square root removing the square. For ${\left({x}^{2}\right)}^{2}$ you remove the outer square.

$3 \times {x}^{2} \times \sqrt{1}$

But $\sqrt{1} = 1$ so we have

$3 \times {x}^{2} \times 1 \text{ "=" } 3 {x}^{2}$
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This trick of multiply by 1 is used in higher maths when you have
$\sqrt{- 4} \to \sqrt{- 1 \times 4} \to 2 \sqrt{- 1}$