More rationalisation help (1 Viewer)

Shazer2 · Mar 18, 2012

Alright, I'm set with the following.
$z^{2} - \frac{1}{z^{2}} \text{ where } z = 1 + \sqrt{2}$

I substitute it in and get to here:
$= (\left{1 + \sqrt{2}}\right)^{2} - \frac{1}{(\left{1 + \sqrt{2}}\right) ^ {2}}$

I'm not sure where to go from there, so if someone could help me out that would be much appreciated.

nightweaver066 · Mar 18, 2012

Rationalise right term first by multiplying numerator and denominator by $(1 - \sqrt{2})^2$

Expand left term.

Combine both terms together for answer.

Shazer2 · Mar 18, 2012

So like:
$(1 + \sqrt{2})^{2} - \frac{1}{(1 + \sqrt{2})^{2}} \times \frac{(1 - \sqrt{2})^{2}}{(1 - \sqrt{2})^{2}}$

Then what?

nightweaver066 · Mar 18, 2012

Shazer2 said:
So like:
$(\left{1 + \sqrt{2}})^{2} - \frac{1}{(\left{1 + \sqrt{2}})^{2}} \times \frac{(\left{1 - \sqrt{2}})^{2}}}{(\left{1 - \sqrt{2}})^{2}}$

Then what?

At least finish off rationalising.. Don't just leave it there..

$(1 + \sqrt{2})^2 - \frac{(1 - \sqrt{2})^2}{(1 + \sqrt{2})^2(1 - \sqrt{2})^2}$

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

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

$= 4\sqrt{2}$

Shazer2 · Mar 18, 2012

Thanks heaps nightweaver066, might bump this thread later for the final question.

More rationalisation help (1 Viewer)

Shazer2

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nightweaver066

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Shazer2

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nightweaver066

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Shazer2

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