# Thread: having algebra issues again

1. ## having algebra issues again

hello,

I had a recent post in which I explained I am new, I am trying to teach myself mathematics. I am having algebra issues again with a limit problem of the indeterminate form.

thank you

2. ## Re: having algebra issues again

Originally Posted by Jonroberts74
I had a recent post in which I explained I am new, I am trying to teach myself mathematics. I am having algebra issues again with a limit problem of the indeterminate form.
You need to learn some algebra first.

Rewrite it as $\frac{{\sqrt {x + 1} - \sqrt {2x + 1} }}{{\sqrt {3x + 4} - \sqrt {2x + 4} }} = - \frac{{\sqrt {3x + 4} + \sqrt {2x + 4} }}{{\sqrt {x + 1} + \sqrt {2x + 1} }}$.

3. ## Re: having algebra issues again

I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you

4. ## Re: having algebra issues again

Originally Posted by Jonroberts74
I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you
You rationalize by multiplying the original expression by
$\left( {\frac{{\sqrt {x + 1} + \sqrt {2x + 1} }}{{\sqrt {3x + 4} + \sqrt {2x + 4} }}\frac{{\sqrt {3x + 4} + \sqrt {2x + 4} }}{{\sqrt {x + 1} + \sqrt {2x + 1} }}} \right)$

Note that you have the "sum and difference" of radicals in the numerator and denominator.

5. ## Re: having algebra issues again

Originally Posted by Jonroberts74
I've been reviewing my algebra books, I couldn't find anything that showed to make it a negative reciprocal. can you explain a little bit? thank you
What Plato showed you is to multiply and divide by the conjugate of the numerator, then multiply and devide by the conjugate of the denominator. Then use the generic formula a^2 - b^2 = (a - b)(a+ b) and simplify. Oh, the conjjugate (a-b) is (a+b).

Replace x by 0 you will get -2.

6. ## Re: having algebra issues again

Thank you very much. I understand now, I don't know why I did not see it before. It can be overwhelming teaching yourself math.

7. ## Re: having algebra issues again

I have another question relating back to this. I spent the last few hours reading about radicals and rationalizing them. In this problem, you multiple numerator and denominator by their respective conjugate then by each others conjugate in order to solve. In another problem I had to rationalize the numerator but not the denominator because it had a radical up top and a rational number on bottom. This left me with a radical on bottom. Previously, I have only rationalized the denominator.

How do I know which Process of the three I need to use? Or for limits do I rationalize each part that contains a radical.

Thank you very much, maybe one day I will be helping people learn math.

9. ## Re: having algebra issues again

I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx

so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h

10. ## Re: having algebra issues again

Originally Posted by Jonroberts74
I ran into another rationalizing problem and it is messing me up. it is f(x) = 1/sqrtx

so when I take my f(a+h) - f(a) I get {[1/sqrt(x+h)] - [1/sqrtx]}/h but inlike the question before that dealt with indeterminate form I am finding rationalizing is getting a bit messy. I end up with [(sqrt(x+h) + sqrtx)/h]/h
It is not really messy. Convert the difference to a common denominator then rationalize the numerator. Take the limit h = 0 and that will be it.