# Thread: Another change the subject question

1. ## Another change the subject question

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

2. Originally Posted by Meggomumsie

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

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

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

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

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

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

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

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

3. Originally Posted by Prove It

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

4. Originally Posted by skeeter
...
Of course...

I didn't want to give EVERYTHING away

5. Originally Posted by skeeter
...
$b > \sqrt{2}$

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

unless the OP is including the complex numbers

6. Thanks for your answer I understand what you have done, but I don't understand how we get from

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

to

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

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

$b^2 - 2 \ge 0$

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

hence the need for the $\pm$

8. Originally Posted by Meggomumsie
Thanks for your answer I understand what you have done, but I don't understand how we get from

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

to

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

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

9. Originally Posted by Meggomumsie
Thanks for your answer I understand what you have done, but I don't understand how we get from

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

to

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

Using a basic index law

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

We have

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

$= 4(b^2 - 2)$.

10. Originally Posted by skeeter
correction ...

$b^2 - 2 \ge 0$

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

hence the need for the $\pm$
How could I forget that