Easy Radical Simplification Exercises for Homework | Quick Tips and Solutions

[MathJakob]

Homework Statement

Simplify: $\sqrt{\frac{6}{81y^6}}$

The Attempt at a Solution

If it was just $\sqrt{\frac{6}{81y}}$ I can just say $\frac{\sqrt{6}}{9\sqrt{y}}$ but the power 6 is throwing me off.

Just a couple more to save me making more topics. $\sqrt{\frac{a^5}{2}}$ can't be simplified?

And... $\sqrt[4]{\frac{2a^8}{b^2c^3}}$ I know I need to multiply it by something but I just can't see it.

Thank you.

 

[SteamKing]

If a^3 * a^3 = a^6, then what is SQRT (a^6)?

 

[MathJakob]

If a^3 * a^3 = a^6, then what is SQRT (a^6)?

$\sqrt{a^6}=a^3$

Thanks. What about the others? Any comment for them?

 

[QuantumCurt]

Just a couple more to save me making more topics. $\sqrt{\frac{a^5}{2}}$ can't be simplified?

This one is a similar idea to the last one. The $$a^5$$ can be broken apart into some factors that are perfect squares. For instance, $$a^4*a$$

From that point then, can you simplify it anymore?

The 2 doesn't have a square root, obviously. Do they ask you to rationalize the denominator?


With the more complicated one, remember that any even root can be seen in terms of square roots. Recall that if you have $$\sqrt(a^2)$$, the root and the power cancel each other.

 

[Ray Vickson]

$\sqrt{a^6}=a^3$

Thanks. What about the others? Any comment for them?

Be careful here: if $a \geq 0$ then $\sqrt{a^6} = a^3,$ but if $a < 0$ then $\sqrt{a^6} = -a^3 \: (> 0!).$ The best way to handle both cases at once is to write $\sqrt{a^6} = |a|^3.$

Note: the symbol $\sqrt{}$ customarily means the non-negative root; if we want a negative root we write $-\sqrt{}$.

 

[MathJakob]

This one is a similar idea to the last one. The $$a^5$$ can be broken apart into some factors that are perfect squares. For instance, $$a^4*a$$

From that point then, can you simplify it anymore?

$\sqrt{\frac{a^5}{2}}=\sqrt{\frac{a^4\times a}{2}}$ = $\sqrt{\frac{a^2\cdot a}{2}}$ = $\frac{a^3}{\sqrt{2}}$ is this better?

Do they ask you to rationalize the denominator?

No

 

[symbolipoint]

Read posts #4 and #5 carefully. The root index is 2, so every two occurrence of the same factor can be put to the outside of the radical symbol as ONE occurrence.

$\sqrt{\frac{a^5}{2}}$
$\sqrt{\frac{a^2a^2a}{2}}$
$a^{2}\sqrt{\frac{a}{2}}$

and then you want to rationalize the denominator.

 

[QuantumCurt]

$\sqrt{\frac{a^5}{2}}=\sqrt{\frac{a^4\times a}{2}}$ = $\sqrt{\frac{a^2\cdot a}{2}}$ = $\frac{a^3}{\sqrt{2}}$ is this better?

That's close. We broke apart the a^5 though so that would find factors of it that are an even power. Remember, when taking the square root of an even numbered exponent, you can just cut the exponent in half, and you'll have the square root. So, when we break the a^5 apart into a^4*a, we can simplify the a^4.

$\sqrt{\frac{a^5}{2}}=\sqrt{\frac{a^4\times a}{2}}$ = $\sqrt{\frac{a^4\cdot a}{2}}$ = $a^2\frac{\sqrt{a}}{\sqrt{2}}$

Which is the same thing as $a^{2}\sqrt{\frac{a}{2}}$

Remember that when you take the square root of something, it goes out in front of the radical. If there is nothing left inside of the radical after taking the root of something, then the radical goes away.

 

[MathJakob]

Sinse this book I'm working from doesn't have answers I'm relying on Wolfram but sometimes it goes a bit weird. Is this right because wolfram is telling me I'm wrong (again)

$\sqrt{\frac{3}{5b^5}} = \sqrt{\frac{3}{5b^2b^2b}} = b^2\sqrt{\frac{3}{5b}}$ ?

Wolfram tells me the answer is $\sqrt{\frac{3}{5}} \sqrt{\frac{1}{a^5}}$

and this one: $\sqrt{\frac{15}{8b^3}} = \sqrt{\frac{15}{8b\cdot b\cdot b}} = b\sqrt{\frac{15}{8b}}$ ?

again wolfram tells me the answer is $\sqrt{\frac{1}{2}}\sqrt{\frac{15}{2}}\sqrt{\frac{1}{b^3}}$

Sorry for keep adding extra questions lol but it stops me spamming new topics.

$\sqrt{\frac{9}{50xy^7}} = \frac{3}{5y^3\sqrt{2xy}}$ is this correct?

 

[QuantumCurt]

$\sqrt{\frac{3}{5b^5}} = \sqrt{\frac{3}{5b^2b^2b}} = b^2\sqrt{\frac{3}{5b}}$ ?

Wolfram tells me the answer is $\sqrt{\frac{3}{5}} \sqrt{\frac{1}{a^5}}$

You got it right. Wolfram did that in a weird way. You can certainly factor and separate what's inside the radical like Wolfram did. $\sqrt{\frac{b}{ax}}$ is the same thing as $\sqrt{\frac{b}{a}}\sqrt{\frac{1}{x}}$ Though I don't know why Wolfram didn't simplify the $b^5$


edit- Whoops, I wasn't paying attention. You're close, but remember that the variable was in the denominator of the expression within the radical.

That would give you-

$$\frac{\sqrt{3}}{b^2\sqrt{5b}}$$

Or you could separate it into-

$$\frac{1}{b^2} \sqrt{\frac{3}{5b}}$$

 

[QuantumCurt]

and this one: $\sqrt{\frac{15}{8b^3}} = \sqrt{\frac{15}{8b\cdot b\cdot b}} = b\sqrt{\frac{15}{8b}}$ ?

again wolfram tells me the answer is $\sqrt{\frac{1}{2}}\sqrt{\frac{15}{2}}\sqrt{\frac{1}{b^3}}$

Wolfram is simplifying these in really unusual ways. I assume you're being asked to simplify this by removing the perfect squares from beneath the radical. You're close with this one.

$$b\sqrt{\frac{15}{8b}}$$

What can you still do with that 8? Look at the factors of 8 and see if you can simplify it some more. Remember that the b was in the denominator of the expression under the radical as well.

$\sqrt{\frac{9}{50xy^7}} = \frac{3}{5y^3\sqrt{2xy}}$ is this correct?

That is indeed correct.

Are you familiar with rationalizing the denominator? Algebra teachers often want you to remove the radical from the denominator.

 

[MathJakob]

Wolfram is simplifying these in really unusual ways. I assume you're being asked to simplify this by removing the perfect squares from beneath the radical. You're close with this one.

$$b\sqrt{\frac{15}{8b}}$$

What can you still do with that 8? Look at the factors of 8 and see if you can simplify it some more. Remember that the b was in the denominator of the expression under the radical as well.



That is indeed correct.

Are you familiar with rationalizing the denominator? Algebra teachers often want you to remove the radical from the denominator.

Nah, the book hasn't mentioned anything about rationalising the denominator. Could you explain?

 

[QuantumCurt]

Nah, the book hasn't mentioned anything about rationalising the denominator. Could you explain?

When you're rationalizing the denominator, the goal is to remove the radical from the denominator. Since square roots of non-perfect squares result in irrational numbers, or infinite decimals that don't repeat, the process is called rationalizing. To do so, you multiply both top and bottom of the expression by the radical that is in the denominator.

$$\frac{\sqrt{3}}{b^2\sqrt{5b}}$$

I'll use this one as an example.

$$\frac{\sqrt{3}}{b^2\sqrt{5b}}$$
$$\frac{\sqrt{3}}{b^2\sqrt{5b}}\frac{\sqrt{5b}}{\sqrt{5b}}$$

Note that we're multiplying the expression by $\frac{\sqrt{5b}}{\sqrt{5b}}$ which is the same thing as multiplying it by 1, which does not change the value of the expression. We're just multiplying it by 1 in such a way that it changes the way the expression is written.

Then we multiply both top and bottom-

$$\frac{\sqrt{3}}{b^2\sqrt{5b}}\frac{\sqrt{5b}}{\sqrt{5b}}$$

$$\frac{\sqrt{3}\sqrt{5b}}{b^2\sqrt{5b}\sqrt{5b}}$$


Then we continue to simplify-

$$\frac{\sqrt{15b}}{b^2\sqrt{5^2b^2}}$$

Then simplify the contents of the radical in the denominator-

$$\frac{\sqrt{15b}}{b^2(5b)}$$

$$\frac{\sqrt{15b}}{5b^3}$$

Now you have a rational denominator. This resultant expression is exactly the same value as the original expression, we've just written in such a way that the denominator has been rationalized. The importance of doing this is somewhat negligible in many fields of math, but algebra teachers will often ask you to do it.