How to rationalise this denominator? (1 Viewer)

Shazer2 · Mar 15, 2012

I'm just wondering how to rationalise the following equation (I tried making the denominator the same and multiplying the numerator by whatever I did to the bottom, but I still got the wrong answer):

$\sqrt{3}-2+\frac{1}{\sqrt{3}-2}$

Shazer2 · Mar 15, 2012

qrpw, will it just be:
$\sqrt{3}-2+\frac{1}{\sqrt{3}-2} \times \frac{\sqrt{3}+2}{\sqrt{3}+2}$

Is that right?

Drongoski · Mar 15, 2012

Shazer2 said:
I'm just wondering how to rationalise the following equation (I tried making the denominator the same and multiplying the numerator by whatever I did to the bottom, but I still got the wrong answer):

$\sqrt{3}-2+\frac{1}{\sqrt{3}-2}$

$= \sqrt 3 - 2 + \frac {1}{\sqrt 3 - 2} \times \frac {\sqrt 3 + 2}{\sqrt 3 + 2} = \sqrt 3 - 2 + \frac {\sqrt 3 + 2}{(\sqrt 3)^2 - 2^2} = \sqrt 3 - 2 + \frac {\sqrt 3 + 2}{3-4}$

$= \sqrt 3 - 2 - \sqrt 3 - 2 = -4$

Carrotsticks · Mar 15, 2012

Shazer2 said:
I'm just wondering how to rationalise the following equation (I tried making the denominator the same and multiplying the numerator by whatever I did to the bottom, but I still got the wrong answer):

$\sqrt{3}-2+\frac{1}{\sqrt{3}-2}$

Alternatively:

$\\ \sqrt{3}-2+\frac{1}{\sqrt{3}-2} \\\\ = \frac{(\sqrt3 - 2)^2 + 1}{\sqrt 3 - 2} \\\\ = \frac{-4 (\sqrt 3 - 2)}{\sqrt 3 - 2} \\\\ = -4$

Shazer2 · Mar 15, 2012

Thanks heaps Drongoski and everyone else.

How to rationalise this denominator? (1 Viewer)

Shazer2

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