Multiplying radicals which include a variable?

Question

What is the $\sqrt {2a}\sqrt {2a}$, $\sqrt{4a}$? Thanks! I know obviously $\sqrt{2}\sqrt{2}$ is $\sqrt{4}$, which is $2$, but what if there is a variable involved?

Answer 1

Assuming $a$ is positive,

$\sqrt{2a}×\sqrt{2a}=\sqrt{2a×2a}=\sqrt{4a^2}=\sqrt4×\sqrt{a^2}=2a$.

In general, multiplying the square root of a number, by the same square root, yields that number, because it is the same as squaring the square root of a number.

If $a$ is negative, let $b=-a$. Consider that $b$ is positive.

$\sqrt{2a}×\sqrt{2a}\\=\sqrt{2b×(-1)}×\sqrt{2b×(-1)}\\=\sqrt{2b}×i×\sqrt{2b}×i\\=\sqrt{2b}×\sqrt{2b}×i×i\\=2b×(-1)\\=-2b\\=2a$.

Answer 2

We have $$\sqrt{2a}\times\sqrt{2a}=(2a)^{\frac12}\times(2a)^{\frac12}=(2a)^{\frac12+\frac12}=2a.$$

Answer 3

The square and the square root are inverses of each other so that

$$\sqrt x\sqrt x=\left(\sqrt x\right)^2=x$$ whateve $x$ can be (provided it is non-negative).

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We also have

$$\sqrt{x\cdot x}=\sqrt{x^2}=x.$$

Answer 4

To see weather or not $\sqrt{2a} \sqrt {2a}$ is the same as $\sqrt {4a}$, we just have to check wether or not $(\sqrt{2a} \sqrt {2a}) \cdot (\sqrt{2a} \sqrt {2a})=4a$.

But $(\sqrt{2a} \sqrt {2a}) \cdot (\sqrt{2a} \sqrt {2a})=(2a) \cdot (2a)=4a^2$

So $\sqrt{2a} \sqrt {2a}$ is not the same as $\sqrt {4a}$.