1
$\begingroup$

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!

$\endgroup$
  • 3
    $\begingroup$ No, rather $\sqrt{a\times a\times a\times a\times a\times a}=\sqrt{(a\times a\times a)^2}=a\times a\times a$. $\endgroup$ – Daniel R Sep 13 '16 at 7:47
  • 5
    $\begingroup$ 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 $a<0!$ $\endgroup$ – gammatester Sep 13 '16 at 7:48
  • $\begingroup$ 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. $\endgroup$ – Kyle Miller Sep 14 '16 at 6:39
14
$\begingroup$

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

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

$\endgroup$
4
$\begingroup$

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

Incorrect.

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

====

----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 $a \ge 0$.]

=====

----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:

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

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

$\endgroup$
  • 1
    $\begingroup$ I do not know why you post this wrong answer after my comment and the correct answer: $8=\sqrt{64} = \sqrt{(-2)^6} \ne (-2)^3 = -8$ $\endgroup$ – gammatester Sep 13 '16 at 8:27
  • $\begingroup$ 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 $a \ge 0$. $\endgroup$ – fleablood Sep 13 '16 at 15:05
2
$\begingroup$

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, $(9)^\frac{1}{2} = (3^2)^\frac{1}{2}=3^{{2}\times\frac{1}{2}}=3$.

$\endgroup$
  • 4
    $\begingroup$ No, $\sqrt{9} = 3$ is correct. When taking square roots of positive numbers, the symbol $\sqrt{x}$ represents the positive square root only. $\endgroup$ – B. Goddard Sep 13 '16 at 12:17
  • $\begingroup$ 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. :) $\endgroup$ – MonK Sep 13 '16 at 12:42
  • 5
    $\begingroup$ 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 $\pm\sqrt{9}$. $\endgroup$ – B. Goddard Sep 13 '16 at 12:45
  • $\begingroup$ "√≠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 $\sqrt{a} = \pm z$. $\endgroup$ – fleablood Sep 13 '16 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.