# Zeros of this equation

• Feb 22nd 2011, 01:12 PM
youngb11
Zeros of this equation
$\displaystyle \displaystyle f(x)=4(x-3)^{\frac{2}{3}}-2(x-3)^{\frac{4}{3}}$

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

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

Edit: Opps, I made a typo.
• Feb 22nd 2011, 01:31 PM
pickslides
Maybe $\displaystyle 3-2\sqrt{2}$
• Feb 22nd 2011, 01:36 PM
Quote:

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

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

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

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

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

the zeros are

$\displaystyle u=0\Rightarrow\ x=3$

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

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

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

$\displaystyle x=3\pm2\sqrt{2}$
• Feb 22nd 2011, 01:59 PM
Soroban
Hello, youngb11!

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

Quote:

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

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

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

And we have two equations to solve.

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

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

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

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

Ah, Archie beat me to it!
• Feb 22nd 2011, 02:00 PM
youngb11
Quote:

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

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

the zeros are

$\displaystyle u=0\Rightarrow\ x=3$

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

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

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

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

Thanks a lot! I overlooked the 2/3 exponent.