Irrational Equations

**Logic** · October 25th 2009, 03:55 PM

"I seem to have reached an odd functional impasse. I am stuck. "

On both.

**Soroban** · October 25th 2009, 04:59 PM

Hello, Logic!

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

Here's the first one . . .

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

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

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

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

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

And we have two equations to solve:

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

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

**Logic** · October 26th 2009, 02:46 AM

Thanks!
This was extremely helpful.

However, the second one still greatly troubles me. I have several more equations of the like, but none seem to be... inspirational enough for me to come up with a common method for solving them.

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