factorising quadratic formulas

**pleasehelpme** · November 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.
Can someone please help me?
Thanks!

**mr fantastic** · November 12th 2008, 12:46 AM

Originally Posted by pleasehelpme

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.
Can someone please help me?
Thanks!

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

and the coup de grace is left for you.

**pleasehelpme** · November 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?

**mr fantastic** · November 12th 2008, 01:17 AM

Originally Posted by pleasehelpme

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:

Originally Posted by mr fantastic

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

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

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

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

Square both sides:

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

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.

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