You rationalize by multiplying the original expression by
$\displaystyle \left( {\frac{{\sqrt {x + 1} + \sqrt {2x + 1} }}{{\sqrt {3x + 4} + \sqrt {2x + 4} }}\frac{{\sqrt {3x + 4} + \sqrt {2x + 4} }}{{\sqrt {x + 1} + \sqrt {2x + 1} }}} \right)$
Note that you have the "sum and difference" of radicals in the numerator and denominator.
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.
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.
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