# square root in algebra

Have a question about roots.

$$\sqrt{9} = 3^2 = 3 \times 3$$ $$\sqrt{9a^6} = 3a^3$$

But why does $\sqrt{a^6} = a^3$?

Is it correct vision?

$$\sqrt{a^6} = \sqrt{a\times a \times a \times a \times a \times a} = \sqrt{2(a \times a \times a)}$$ Can anybody add more details about my question?

Thanks!

• No, rather $\sqrt{a\times a\times a\times a\times a\times a}=\sqrt{(a\times a\times a)^2}=a\times a\times a$. – Daniel R Sep 13 '16 at 7:47
• This is not generally correct. The results of the square root function are $\ge 0$ and $a^6 \ge 0$. So your equation breaks down for $a<0!$ – gammatester Sep 13 '16 at 7:48
• You can always test particular values of your variables. For instance, with $a=2$, $\sqrt{2^6}=\sqrt{64}=8$, but $\sqrt{2(2\cdot 2\cdot 2)}=\sqrt{16}=4$. And, $8=2^3$, as expected. – Kyle Miller Sep 14 '16 at 6:39

## 3 Answers

It is not a correct vision, the correct one would be:

$$\sqrt{a^6}=\sqrt{a\times a\times a\times a\times a\times a}=\sqrt{(a\times a\times a)^2}=\vert a\times a\times a \vert=\vert a^3\vert.$$

-----$\sqrt{9} = 3^2 = 3 * 3$

Incorrect.

$\sqrt{9} = \sqrt{3^2} = 3$

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----but why $\sqrt{a^6}=a^3$

Because $\sqrt{a^6} = \sqrt{(a^3)^2} = a^3$.

[Assuming $a \ge 0$. If $a < 0$ then $a^6 = |a|^6 =|a^6| > 0$ and $\sqrt{a^6} = |a|^3$. For the rest of the post I'm assuming $a \ge 0$.]

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----Is it correct vision ? $\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{2(a∗a∗a)}$

Not quite. $\sqrt{a^6} = \sqrt{a∗a∗a∗a∗a∗a}=\sqrt{(a∗a∗a)*(a*a*a)}=\sqrt{a^3*a^3} = \sqrt{(a^3)^2} = a^3$

Note:

$(a*a*a*a*a*a) \ne 2(a*a*a)$

$(a*a*a*a*a*a) = (a*a*a)*(a*a*a) = (a*a*a)^2$.

• I do not know why you post this wrong answer after my comment and the correct answer: $8=\sqrt{64} = \sqrt{(-2)^6} \ne (-2)^3 = -8$ – gammatester Sep 13 '16 at 8:27
• I'm teaching to the audience and answering the OP's specific question. The OP has been given an answer sheet that states $\sqrt{9a^6} = 3a^3$. As such I must conclude the answer sheet has specifically stated $a \ge 0$. – fleablood Sep 13 '16 at 15:05

In your vision, you grouped $a*a*a$ together and multiplied. But you forgot that, $x*x=x^2\text{ and}\neq 2x$.

So, even in your groupings

$(a*a*a)*(a*a*a)=(a*a*a)^2$

To your originial question, If you are ever in doubt remember, $\sqrt x$ is same as saying $x^\frac{1}{2}$.

So, if you put $x=9a^6$ you will get, $(9a^6)^\frac{1}{2}$.

The exponent will distribute itself over the bases, using this identity $(xy)^n=x^n\times y^n$.

So overall, the expression will reduce to $(9)^\frac{1}{2}\times(a^6)^\frac{1}{2}$

Now, comes a third identity $(x^m)^n=x^{m\times n}$. So, $(9)^\frac{1}{2} = (3^2)^\frac{1}{2}=3^{{2}\times\frac{1}{2}}=3$.

• No, $\sqrt{9} = 3$ is correct. When taking square roots of positive numbers, the symbol $\sqrt{x}$ represents the positive square root only. – B. Goddard Sep 13 '16 at 12:17
• I think your interpretation is slightly incorrect. Just to be sure, a quick google search reveals this ..."Every positive number a has two square roots: √a, which is positive, and −√a, which is negative." You might want to try that. :) – MonK Sep 13 '16 at 12:42
• Yes, every positive number has two square roots. But the symbol $\sqrt{9}$ represents the positive square root. If you want to write both square roots, you need to write $\pm\sqrt{9}$. – B. Goddard Sep 13 '16 at 12:45
• "√≠3 rather it is ±3, because (−3)^2=(3)^2=9" and ."Every positive number a has two square roots: √a, which is positive, and −√a, which is negative." Those are contradictory statements. $\sqrt{a}$ can't be positive if $\sqrt{a} = \pm z$. – fleablood Sep 13 '16 at 15:17