# Irrational Equations

• Oct 25th 2009, 03:55 PM
Logic
Irrational Equations
"I seem to have reached an odd functional impasse. I am stuck. "

On both.
• Oct 25th 2009, 04:59 PM
Soroban
Hello, Logic!

There are not easy problems.
They required special treatment, delicate handling.

Here's the first one . . .

Quote:

$\displaystyle \sqrt[3]{(x-1)^2} + \sqrt[3]{(x-2)^2} \;=\;\tfrac{5}{2}\sqrt[3]{x-2}\sqrt[3]{x-2}$

We have: .$\displaystyle (x-1)^{\frac{2}{3}} + (x-2)^{\frac{2}{3}} \;=\;\tfrac{5}{2}(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}}$

. . . . . . $\displaystyle 2(x-1)^{\frac{2}{3}} + 2(x-2)^{\frac{2}{3}} \;=\;5(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}}$

. . . . . . $\displaystyle 2(x-1)^{\frac{2}{3}} - 5(x-1)^{\frac{1}{3}}(x-2)^{\frac{1}{3}} + 2(x-2)^{\frac{2}{3}} \;=\;0$

Factor: .$\displaystyle \bigg[2(x-1)^{\frac{1}{3}} - (x-2)^{\frac{1}{3}}\bigg]\,\bigg[(x-1)^{\frac{1}{3}} - 2(x-2)^{\frac{1}{3}}\bigg] \;=\;0$

And we have two equations to solve:

$\displaystyle 2(x-1)^{\frac{1}{3}} - (x-2)^{\frac{1}{3}} \;=\;0 \quad\Rightarrow\quad 2(x-1)^{\frac{1}{3}} \;=\;(x-2)^{\frac{1}{3}}$
. . Cube both sides: .$\displaystyle 8(x-1) \;=\;x-2 \quad\Rightarrow\quad 8x-8 \:=\:x-2 \quad\Rightarrow\quad 7x\:=\:6 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{6}{7}}$

$\displaystyle (x-1)^{\frac{1}{3}} - 2(x-2)^{\frac{1}{3}} \:=\:0 \quad\Rightarrow\quad 2(x-2)^{\frac{1}{3}} \:=\:(x-1)^{\frac{1}{3}}$
. . Cube both sides: .$\displaystyle 8(x-2) \:=\:x-1 \quad\Rightarrow\quad 8x - 16 \:=\:x-1 \quad\Rightarrow\quad 7x \:=\:15 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{15}{7}}$

• Oct 26th 2009, 02:46 AM
Logic
Thanks!