$$\frac{a}{a-\sqrt{a^2-16}}$$ is the expression. I am not sure if i answered it right but please help me do this.
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Daniel Fischer
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ikaidubidu
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Is $$\frac{a}{a-\sqrt{a^2-16}}$$ the correct interpretation of the expression? – Daniel Fischer Aug 18 '13 at 14:55
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@DanielFischer yes. – ikaidubidu Aug 18 '13 at 14:56
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Where is the equation? – lab bhattacharjee Aug 18 '13 at 14:56
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@labbhattacharjee \frac{a}{a-\sqrt{a^2-16}} – ikaidubidu Aug 18 '13 at 14:57
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@ikaidubidu, the only thing we can do is rationalize the denominator $$\frac a{a-\sqrt{a^2-16}}=\frac{a(a+\sqrt{a^2-16})}{a^2-(a^2-16)}=\cdots$$. – lab bhattacharjee Aug 18 '13 at 15:00
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What you have is an expression, but not an equation. We can indeed simplify the expression.
We can multiply by the conjugate of the denominator, to obtain a difference of squares., e.g. $$\begin{align}\frac{a}{a-\sqrt{a^2-16}} & = \frac{a(a + \sqrt{a^2 - 16)}}{(a - \sqrt {a^2 - 16})(a + \sqrt{a^2 - 16})} \\ \\ & = \frac{a(a + \sqrt{a^2 - 16)}}{a^2 - (a^2 - 16)}\\ \\ & = \frac{{a(a + \sqrt{a^2 - 16)}}}{ 16}\end{align}$$
Added: What we've accomplished is called "rationalizing the denominator", which simply means getting the "radical" out of the denominator.
amWhy
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