# Solving Equations Quadratic In Form (By Substitution?)

• September 5th 2011, 07:22 AM
street1030
Solving Equations Quadratic In Form (By Substitution?)
I know how to solve by substituting in some instances. The only times I've done it is with examples like this (x +1)^2/3 = (x + 1)^1/3

With this problem since (x+1)^2/3 is [(x+1)^1/3]^2 you can just substitute "u" for the whole thing leaving you with u^2 = u. You would then bring everything to one side and solve the equation from there.

The problem I'm having trouble with though looks a bit different.

(x-3)^2/5 = (4x)^1/5

Now the exponent 2/5 is the square of the exponent 1/5 but x-3 & 4x don't match so you shouldn't be able to just replace them with u.

Anyone know where I should start?
• September 5th 2011, 07:59 AM
Soroban
Re: Solving Equations Quadratic In Form (By Substitution?)
Hello, street1030!

A sneaky one! . . . It doesn't involve substitution.

Quote:

$(x-3)^{\frac{2}{5}} \:=\: (4x)^{\frac{1}{5}}$

Raise both sides to the 5th power: . $(x-3)^2 \:=\:4x$

And we have: . $x^2 - 6x + 9 \:=\:4x \quad\Rightarrow\quad x^2 - 10x + 9 \:=\:0$

Got it?
• September 5th 2011, 08:13 AM
street1030
Re: Solving Equations Quadratic In Form (By Substitution?)
AH! Didn't think of that. Thanks a bunch. Think this will help me with a few more also.