Question

how can i prove that:
(square(x) - square(a))/(x-a) = 1/(square(x)+square(a))

Answer 1

It doesn't. To let the first part be true, x - a does not equal 0 ie\[x \neq a\]
But x^2 - a^2 = (x + a)(x - a) through factoring, so
\[\frac{ x^2 - a^2 }{ x - a } = \frac{ (x + a)(x - a) }{ x - a } = x+a\]
and\[x + a \neq \frac { 1 } { x^2 - a^2 }\]
Did I read your question correctly? :/

Answer 2

Hi misha511, thank you for your answer.
I think you made a mistake or I did. when I wrote square(x)- square(a) I mean x^(1/2) - a^(1/2). Sorry, but my english is not very good. I tried to use the notation I use in Excel.

Answer 3

Oh OK. "sqrt" means square root. I thought about that, and decided against it. One sec

Answer 4

So then it follows the same principle:
\[a^2 - b^2 = (a+b)(a-b)\]
Therefore:
\[\frac{ \sqrt{x} - \sqrt{a} }{ x - a } = \frac{ \sqrt{x} - \sqrt{a} }{ (\sqrt{x} + \sqrt{a})(\sqrt{x} - \sqrt{a}) }\]
Then divide top and bottom by sqrt(x) - sqrt(a)
\[\frac {\frac{ \sqrt{x} - \sqrt{a} }{\sqrt{x} - \sqrt{a}}}{ \frac{(\sqrt{x} + \sqrt{a})(\sqrt{x} - \sqrt{a})}{\sqrt{x} - \sqrt{a}} }\]
The top cancels leaving 1, and the bottom partially cancels, leaving \[\sqrt{x} + \sqrt{a}\]
And therefore we're left with \[\frac{1}{\sqrt{x} - \sqrt{a}}\]

Answer 5

misha511, thank you for your help.