• Nov 12th 2008, 12:34 AM
Im having trouble factorising:

(x/(16 + x²)^½) - 1/3 = 0

Thats the square root of (16+x*x) divided by x, minus 1/3. I have to find the values that make it equal 0.
Thanks!
• Nov 12th 2008, 12:46 AM
mr fantastic
Quote:

Im having trouble factorising:

(x/(16 + x² )^½) - 1/3 = 0

Thats the square root of (16+x*x) divided by x, minus 1/3. I have to find the values that make it equal 0.
Thanks!

$\displaystyle \frac{x}{\sqrt{16 + x^2}} - \frac{1}{3} = 0 \Rightarrow x = \frac{\sqrt{16 + x^2}}{3} \Rightarrow x^2 = \frac{16 + x^2}{9} \Rightarrow 9x^2 = 16 + x^2$

and the coup de grace is left for you.
• Nov 12th 2008, 01:13 AM
How did you get the equation over 9 in step 3? and how did you get the equation to equal x-squared?
• Nov 12th 2008, 01:17 AM
mr fantastic
Quote:

How did you get the equation over 9 in step 3? and how did you get the equation to equal x-squared?

Mr F edits:

Quote:

Originally Posted by mr fantastic
$\displaystyle \frac{x}{\sqrt{16 + x^2}} - \frac{1}{3} = 0$

$\displaystyle {\color{red}\Rightarrow \frac{x}{\sqrt{16 + x^2}} = \frac{1}{3}}$

Multiply both sides by $\displaystyle {\color{red}\sqrt{16 + x^2}}$:

$\displaystyle \Rightarrow x = \frac{\sqrt{16 + x^2}}{3}$

Square both sides:

$\displaystyle \Rightarrow x^2 = \frac{16 + x^2}{9} \Rightarrow 9x^2 = 16 + x^2$

and the coup de grace is left for you.

If you're attempting these sorts of questions you're expected to be comfortable with this sort of algebraic manipulation.