# Another change the subject question

Printable View

• Sep 27th 2009, 08:13 AM
Meggomumsie
Another 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 AM
Prove It
Quote:

Originally Posted by Meggomumsie
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.

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

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

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

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

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

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

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

$\displaystyle b = \frac{\sqrt{a^2 + 8}}{2}$.
• Sep 27th 2009, 08:20 AM
skeeter
Quote:

Originally Posted by Prove It

$\displaystyle b = \textcolor{red}{\pm}\frac{\sqrt{a^2 + 8}}{2}$.

...
• Sep 27th 2009, 08:22 AM
Prove It
Quote:

Originally Posted by skeeter
...

Of course...

I didn't want to give EVERYTHING away (Giggle)
• Sep 27th 2009, 08:23 AM
e^(i*pi)
Quote:

Originally Posted by skeeter
...

$\displaystyle b > \sqrt{2}$

$\displaystyle a \geq 0$ and since $\displaystyle a^2 +8 > 0$ the negative solution can be discarded

unless the OP is including the complex numbers
• Sep 27th 2009, 08:29 AM
Meggomumsie
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 AM
skeeter
Quote:

Originally Posted by e^(i*pi)
$\displaystyle b > \sqrt{2}$

correction ...

$\displaystyle b^2 - 2 \ge 0$

$\displaystyle b \ge \sqrt{2}$ or $\displaystyle b \le -\sqrt{2}$

hence the need for the $\displaystyle \pm$
• Sep 27th 2009, 08:34 AM
skeeter
Quote:

Originally Posted by Meggomumsie
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?

$\displaystyle (2\sqrt{b^2-2})^2 = 2^2 \cdot (\sqrt{b^2-2})^2 = ?$
• Sep 27th 2009, 08:34 AM
Prove It
Quote:

Originally Posted by Meggomumsie
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?

Using a basic index law

$\displaystyle (a\cdot b)^n = a^n\cdot b^n$

We have

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

$\displaystyle = 4(b^2 - 2)$.
• Sep 27th 2009, 08:40 AM
e^(i*pi)
Quote:

Originally Posted by skeeter
correction ...

$\displaystyle b^2 - 2 \ge 0$

$\displaystyle b \ge \sqrt{2}$ or $\displaystyle b \le -\sqrt{2}$

hence the need for the $\displaystyle \pm$

How could I forget that (Sleepy)