# Thread: Zeros of this equation

1. ## Zeros of this equation

$\displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}$

Working it out, I got two x-intercepts of $\displaystyle x=3$ and $\displaystyle x=3+2\sqrt{2}$

However, there is apparently a third root. What am I missing:S?

Edit: Opps, I made a typo.

2. Maybe $3-2\sqrt{2}$

3. Originally Posted by youngb11
$\displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}$

Working it out, I got two x-intercepts of $\displaystyle x=3$ and $\displaystyle x=1+2\sqrt{2}$

However, there is apparently a third root. What am I missing:S?
$\displaystyle\ u=(x-3)^{\frac{2}{3}}$

$4u-2u^2=2u(2-u)$

the zeros are

$u=0\Rightarrow\ x=3$

$\displaystyle\ u=2\Rightarrow\ (x-3)^{\frac{2}{3}}=2$

$\Rightarrow\ (x-3)^2=2^3=8$

$\Rightarrow\ x-3=\pm\sqrt{8}=\pm2\sqrt{2}$

$x=3\pm2\sqrt{2}$

4. Hello, youngb11!

You took a square root . . . but it's easy to overlook.

$\displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}$

$\text{We have: }\;4(x-3)^{\frac{2}{3}} - 2(x-3)^{\frac{4}{3}} \;=\;0$

$\text{Factor: }\;2(x-3)^{\frac{2}{3}}\,\bigg[2 - (x-3)^{\frac{2}{3}}\bigg] \;=\;0$

And we have two equations to solve.

$\text{The first equation is: }\;2(x-3)^{\frac{2}{3}} \:=\:0 \quad\Rightarrow\quad\boxed{x \:=\:3}$

$\text{The second equation is: }\;2 - (x-3)^{\frac{2}{3}} \:=\: 0 \quad \Rightarrow\quad (x-3)^{\frac{2}{3}} \:=\:2$

$\text{Raise both sides to the }\tfrac{3}{2}\text{ power.}$
. . $\text{This means }cube\text{ both sides: }\;(x-3)^2 \;=\;8$
. . $\text{and take the }square\:root\text{ of both sides: }\;x-3 \;=\;\pm\sqrt{8}$

$\text{Hence: }\;\boxed{x \:=\:3 \pm2\sqrt{2}}$

Ah, Archie beat me to it!

5. Originally Posted by Archie Meade
$\displaystyle\ u=(x-3)^{\frac{2}{3}}$

$4u-2u^2=2u(2-u)$

the zeros are

$u=0\Rightarrow\ x=3$

$\displaystyle\ u=2\Rightarrow\ (x-3)^{\frac{2}{3}}=2$

$\Rightarrow\ (x-3)^2=2^3=8$

$\Rightarrow\ x-3=\pm\sqrt{8}=\pm2\sqrt{2}$

$x=3\pm2\sqrt{2}$
Thanks a lot! I overlooked the 2/3 exponent.