3
$\begingroup$

Hello everyone how would I simplify the following radicals.

$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$$

I got $$3 \sqrt{64a^4b^7}$$ I know $64$ square root is $8$ and $a^4$ square root is $a^2$

My second question is how would I simplify the following

$$\frac{3}{4}\sqrt{3t^3}$$

I know $ \sqrt[\large4]{3t^3}$ is equal to $(3t)^{\frac{3}{4}}$ so would I multiply by $\frac{1}{4}$

My final question is how would I simplify the following

$$ \sqrt[\large 6]{x^6y^4}$$

Whomever helps Fernando with these questions shall receive his eternal gratitude.

$\endgroup$
5
  • $\begingroup$ On my second question I would like to say it is ((3))/((4 root((3t^(3)) when I said 4 root I mean it like cube root or square root kind of way as in 3 or 2. $\endgroup$ Commented Jan 17, 2013 at 3:35
  • $\begingroup$ I thought the title of this question meant "these algebra questions are radical!" and was subsequently disappointed. $\endgroup$ Commented Jan 17, 2013 at 3:46
  • $\begingroup$ @proximal Same here! But not the "disappointed" part though. $\endgroup$ Commented Jan 17, 2013 at 3:50
  • $\begingroup$ Sorry I did not notice. I guess I should have titled my post Algebra questions involving radicals. $\endgroup$ Commented Jan 17, 2013 at 4:01
  • $\begingroup$ I know for my first question the final answer is 24a^(2)b^(3)square root(b) but how do I reach that. $\endgroup$ Commented Jan 17, 2013 at 4:10

4 Answers 4

3
$\begingroup$

Hint:$$\sqrt[6]{x^6y^4} = \left(x^6y^4\right)^{\frac{1}{6}} = x^\frac{6}{6}y^\frac{4}{6}$$

$\endgroup$
2
  • 1
    $\begingroup$ excellent hint.thanks $\endgroup$ Commented Jan 17, 2013 at 4:02
  • $\begingroup$ It does work as well. +1 $\endgroup$
    – Mikasa
    Commented Jan 17, 2013 at 7:28
2
$\begingroup$

$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}}$

$$3 \sqrt{2a^3b^5} \sqrt{32ab^{2}} = 3 \sqrt{64a^4b^7} = 3\sqrt{64a^4}\sqrt{b^7} = 3\cdot 8 a^2 \sqrt{b^7} = 24a^2 \sqrt{b^7} = 24a^2b^{\large\frac72}\quad\quad\quad\tag{1}$$


"$ \;\sqrt[\large4]{3t^3}\;$ is equal to $\;(3t)^{\frac{3}{4}}\;$"

Not quite:

$$\sqrt[\large 4]{3t^3} \;=\; (3\,t^3)^{\large \frac 14} \;= \;3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}\tag{2}$$

EDIT: to address comment/question below

If your original expression (to simplify) was $\;\dfrac{3}{\sqrt[\large 4]{3t^3}}\;$ then using the simplification above, we have $$\;\dfrac{3}{\sqrt[\large 4]{3t^3}},\;= \;\frac{3}{3^{\large \frac{1}{4}}\,t^{\large\frac{3}{4}}}\;=\;\frac{3^{\large\frac{4}{4}}\cdot 3^{-\large\frac{1}{4}}}{t^{\large\frac{3}{4}}} \;=\; \frac{3^{\large\frac{3}{4}}}{t^{\large\frac{3}{4}}}\;=\;\frac{(3^3)^{\large\frac{1}{4}}}{(t^3)^{\large\frac{1}{4}}} \;=\;\left(\frac{27}{t^3}\right)^{\large\frac{1}{4}} \;= \; \sqrt[\large 4]{\frac{27}{t^3}}$$


My final question is how would I simplify the following: $\quad \sqrt[\large 6]{x^6y^4}\;$?

$$\sqrt[\large 6]{x^6y^4}\; = \;\left(x^6y^4\right)^{\large \frac{1}{6}}\; =\; x^{\large \frac{6}{6}}y^{\large \frac{4}{6}} \;=\; xy^{\large \frac{2}{3}}\; = \;x\sqrt[\large 3]{y^2}\tag{3}$$

$\endgroup$
5
  • $\begingroup$ I see so now I have ((3))/((3^(1/4)t^(3/4)) the problem is the book final answer is 4 root(27)/(t) I have trouble getting that. $\endgroup$ Commented Jan 17, 2013 at 4:08
  • $\begingroup$ @FernandoMartinez: It is good to know that amWhy used the fact that: $\sqrt[m]{x^n}=x^{\frac{n}{m}}$. Nice +1 $\endgroup$
    – Mikasa
    Commented Jan 17, 2013 at 7:27
  • $\begingroup$ @Fernando: I expanded to consider your posted expression as the denominator of the fraction you give in your comment here. $\endgroup$
    – amWhy
    Commented Jan 17, 2013 at 16:18
  • $\begingroup$ @amWhy thanks for your help. $\endgroup$ Commented Jan 17, 2013 at 16:31
  • $\begingroup$ @FernandoMartinez: de nada... Glad to help $\endgroup$
    – amWhy
    Commented Jan 17, 2013 at 16:37
1
$\begingroup$

$3\sqrt{2a^{3}b^{5}}\sqrt{32ab^{2}} = 3(2a^{3}b^{5})^{1/2}(2^{5}ab^{2})^{1/2} = 3(2^{6}a^{4}b^{7})^{1/2} = 3 \cdot 2^{3}a^{2}b^{7/2}$.

$\endgroup$
0
$\begingroup$

To get to the heart of the matter, since these are square roots, you are looking for the largest factor of the radicand that is a perfect square.

For example, with $\sqrt{32ab^2}$ you are looking for the largest factor of $32ab^2$ that is a perfect square. To answer this question, you must first ask yourself, "What are the factors of $32$, $a$, and $b^2$?" Well, the factors of $32$ are $1, 2, 4, 8, 16,$ and $32$. The factors of $a$ are $1$ and $a$. The factors of $b^2$ are $1, b,$ and $b^2$.

To be even more precise, let's do an example. Say we wanted to simply $\sqrt{300ab^4}$. We would want to find the factors of $300$, $a$, and $b^4$ that are perfect squares. The largest factor of $300$ that is a perfect square is $100$. The largest factor of $a$ that is a perfect square is $1$. The largest factor of $b^4$ that is a perfect square is $b^4$ itself. Thus, the largest quantity we can take out of the radical is $100\cdot1\cdot b^4 = 100b^4$. The rest of the procedure goes like this:

$$ \sqrt{300ab^4} = \sqrt{100b^4 \cdot 3a} = \sqrt{100b^4}\sqrt{3a} = 10b^2\sqrt{3a}. $$

Try this process with the square roots you have in your post and see what happens.

As a side note, if you were instead trying to simplify the cube root of a quantity, you would look for the largest factor of the radicand that is a perfect cube, rather than a perfect square. With fourth roots, you'd look for perfect fourth powers, and so on.

$\endgroup$
3
  • $\begingroup$ For square root(32ab^2) I I simplified and got 4bsuqare root(2a) for 3 square root(2a^3b^5) I got 3ba square root(2ab^3) when I simplified. $\endgroup$ Commented Jan 17, 2013 at 4:24
  • $\begingroup$ That's a good start. First, note that you don't have to worry about the $3$ in front of the radical since it cannot be simplified anymore. This doesn't help much, but carrying an extra $3$ through your simplification steps could cost an error or two. Second, note that $\sqrt{b^3}$ is not as simplified as it could be. $\endgroup$
    – JavaMan
    Commented Jan 17, 2013 at 4:26
  • $\begingroup$ @ Fernando which is the correct one for your problem, is it (1) : $3\sqrt{2a^3b^5}$ or is it (2) : $\sqrt[3]{2a^3b^5}$? $\endgroup$
    – adam W
    Commented Jan 17, 2013 at 5:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .