# Thread: Algebra help needed on solving 2x - 3x^(-1/2) + 1 = 0

1. ## Algebra help needed on solving 2x - 3x^(-1/2) + 1 = 0

2x - 3x^(-1/2) + 1 = 0

I know that when x = 1 that y or the function = 0 but I did not figure that out by myself and needed wolfram alpha to tell me. If I were to do it myself, I would have to plug in numbers which is obviously inefficient.

My question is how do I solve it algebraically? What are the steps? I can't factor out x because there is a +1 and I can't move it to the other side because then it's no longer 0 and therefore no longer easy to figure out the values of x.

Any help would be GREATLY appreciated!

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

$\displaystyle 2x-\frac{3}{\sqrt{x}}+1=0$

let $\displaystyle \sqrt{x}=y$

$\displaystyle x=y^2$

$\displaystyle 2y^2-\frac{3}{y}+1=0$

$\displaystyle 2y^3+y-3=0$

$\displaystyle 2y^3-2y^2+2y^2-2y+3y-3=0$

$\displaystyle 2y^2(y-1)+2y(y-1)+3(y-1)=0$

$\displaystyle (y-1)(2y^2+2y+3)=0$

$\displaystyle y=1$

$\displaystyle x=1^2=1$

Thus, $\displaystyle x=1$