# Solving equation with negative exponents

## Homework Statement

2x^(-1/3)-9x^(-1/6)= -10

?

## The Attempt at a Solution

I have tried to factor out x^(-1/6)
x^(-1/6) (2x-9)= -10
I'm not sure thats even right
I have also converted to fractions
1/2x^(1/3)-1/9x^(1/6)= -10
I'm not sure which route to go or if either is right???

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berkeman
Mentor

## Homework Statement

2x^(-1/3)-9x^(-1/6)= -10

?

## The Attempt at a Solution

I have tried to factor out x^(-1/6)
x^(-1/6) (2x-9)= -10
I'm not sure thats even right
I have also converted to fractions
1/2x^(1/3)-1/9x^(1/6)= -10
I'm not sure which route to go or if either is right???
If you write it this way, can you see what you can do to put the left hand side (LHS) over a common denominator in order to proceed?

$$\frac{2}{x^{1/3}} - \frac{9}{x^{1/6}} = -10$$

Mark44
Mentor
And as a further hint, x1/3 = (x1/6)2, so with the right substitution, your equation is quadratic in form.

Yeah always try to notice when you can turn an equation into a quadratic eg.

$x + x^{-1} + A = 0 \Rightarrow x^2+1+Ax= 0$

$e^x + e^{-x} + A = 0 \Rightarrow (e^{x})^2 +1+ Ae^x = 0$

$\cot(x) + \tan(x) + A = 0 \Rightarrow 1 + \tan^2 x +A\tan x= 0$ etc.

substitutions can be helpful aswell, like substitute e^x for y or something.