Not sure where to start with this one (Smirk)

$\displaystyle a=2\sqrt{b^2-2}$ (Make b the subject of the formula)

How do I begin please? I can't see how to square everything at all this time, not even incorrectly.

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- Sep 27th 2009, 08:13 AMMeggomumsieAnother change the subject question
Not sure where to start with this one (Smirk)

$\displaystyle a=2\sqrt{b^2-2}$ (Make b the subject of the formula)

How do I begin please? I can't see how to square everything at all this time, not even incorrectly. - Sep 27th 2009, 08:17 AMProve It
$\displaystyle a = 2\sqrt{b^2 - 2}$

$\displaystyle a^2 = (2\sqrt{b^2 - 2})^2$

$\displaystyle a^2 = 4(b^2 - 2)$

$\displaystyle \frac{a^2}{4} = b^2 - 2$

$\displaystyle \frac{a^2}{4} + 2 = b^2$

$\displaystyle \frac{a^2 + 8}{4} = b^2$

$\displaystyle \sqrt{\frac{a^2 + 8}{4}} = b$

$\displaystyle b = \frac{\sqrt{a^2 + 8}}{2}$. - Sep 27th 2009, 08:20 AMskeeter
- Sep 27th 2009, 08:22 AMProve It
- Sep 27th 2009, 08:23 AMe^(i*pi)
- Sep 27th 2009, 08:29 AMMeggomumsie
Thanks for your answer I understand what you have done, but I don't understand how we get from

$\displaystyle a^2=(2\sqrt{b^2-2})^2$

to

$\displaystyle a^2=4(b^2-2)$

Can you explain please? - Sep 27th 2009, 08:32 AMskeeter
- Sep 27th 2009, 08:34 AMskeeter
- Sep 27th 2009, 08:34 AMProve It
- Sep 27th 2009, 08:40 AMe^(i*pi)