# Math Help - Any kind hearted soul, radical expressions help!!!

2. ## Re: Any kind hearted soul, radical expressions help!!!

Hint- $\sqrt{a} = a^{\frac{1}{2}}$. Also, $\sqrt{a} \sqrt{b} = \sqrt{ab}$.

3. ## Re: Any kind hearted soul, radical expressions help!!!

The second problem gives you a hint. What did you get?

$(x^2 + \sqrt{2}xy + y^2) * (x^2 - \sqrt{2}xy + y^2) = what?$

By analogy then $x^8 + y^8 = what?$

4. ## Re: Any kind hearted soul, radical expressions help!!!

Thanks for the reply! As you can see in my image, for question 53, i cannot find any perfect squares in the square root. I do not know how to advance from there on.

As for question 57, how do I go on from there? I do see some similarities with X^4 and Y^4 in my half-completed answer. However, how do I remove the 2x^2y^2-2xy that u see in my answer ?

5. ## Re: Any kind hearted soul, radical expressions help!!!

$\sqrt{x^{2n}} * \sqrt{x^3y^{3n}} * \sqrt{y^{(n+1)}} = \sqrt{x^{2n} * x^3 * y^{3n} * y^{(n + 1)}}= \sqrt{x^{(2n + 3)} * y^{(4n + 1)}}.$ So far so good.

But $2n + 3 = (2n + 2) + 1 = 2(n + 1) + 1.$ So

$\sqrt{x^{(2n + 3)} * y^{4n + 1}} = \sqrt{x^{\{2(n + 1) + 1\}} * y^{(4n+1)}} = \sqrt{x^{\{2(n + 1)\}} * x^1 * y^{4n} * y^1} = x^{(n + 1)} * y^{2n} * \sqrt{xy}.$

As for the second problem, you did the algebra wrong

$(x^2 + \sqrt{2}xy + y^2)(x^2 - \sqrt{2}xy + y^2) = x^4 -\sqrt{2}x^3y + x^2y^2 + \sqrt{2}x^3y - 2x^2y^2 + \sqrt{2}xy^3 + x^2y^2 -\sqrt{2}xy^3 + y^4 =$

$x^4 -\sqrt{2}x^3y + \sqrt{2}x^3y + x^2y^2 + x^2y^2 - 2x^2y^2 + \sqrt{2}xy^3 -\sqrt{2}xy^3 + y^4 = x^4 + y^4.$

Now think about $x^8 + y^8.$

6. ## Re: Any kind hearted soul, radical expressions help!!!

hey JeffM. Isn't Squareroot2xy multiplied by squareroot2xy = 2xy? as you remove the squares with the square root.

7. ## Re: Any kind hearted soul, radical expressions help!!!

Originally Posted by xwy
hey JeffM. Isn't Squareroot2xy multiplied by squareroot2xy = 2xy? as you remove the squares with the square root.
You are correct that: $\sqrt{2xy} * \sqrt{2xy} = \sqrt{2xy * 2xy} = \sqrt{2^2x^2y^2} = 2xy,\ assuming\ xy \ge 0.$

But if I understood your problem, that is NOT what you are dealing with.

$\sqrt{2}xy * \sqrt{2}xy = \sqrt{2} * xy * \sqrt{2} * xy = \sqrt{2} * \sqrt{2} * x^2y^2 = \sqrt{2 * 2} * x^2y^2 = \sqrt{4} * x^2y^2 = 2x^2y^2.$

If I have led you astray by not understanding (or by making ANOTHER idiotic mistake), I do apologize.

8. ## Re: Any kind hearted soul, radical expressions help!!!

Hey no worries. I really appreciate your help!

9. ## Re: Any kind hearted soul, radical expressions help!!!

This post should be deleted as soon as the preceding post is, but why anyone would pay money for academic help to someone who cannot bother to proof his text for gross errors in English grammar and spelling is beyond me. I guess it is true that a fool and his money are soon parted.

10. ## Re: Any kind hearted soul, radical expressions help!!!

Originally Posted by JeffM
This post should be deleted as soon as the preceding post is, but why anyone would pay money for academic help to someone who cannot bother to proof his text for gross errors in English grammar and spelling is beyond me. I guess it is true that a fool and his money are soon parted.
I agree with you entirely. However responding to such simply gives the spammer a larger chance of someone seeing his/her post, which is the whole idea. Please report the post and either mash or I will get to it as soon as possible.

-Dan