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!
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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! $\displaystyle \frac{x}{\sqrt{16 + x^2}} - \frac{1}{3} = 0 \Rightarrow x = \frac{\sqrt{16 + x^2}}{3} \Rightarrow x^2 = \frac{16 + x^2}{9} \Rightarrow 9x^2 = 16 + x^2$ and the coup de grace is left for you.
How did you get the equation over 9 in step 3? and how did you get the equation to equal x-squared?
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 $\displaystyle \frac{x}{\sqrt{16 + x^2}} - \frac{1}{3} = 0$ $\displaystyle {\color{red}\Rightarrow \frac{x}{\sqrt{16 + x^2}} = \frac{1}{3}}$ 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.
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