Question 70150
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{{{2x - 3x^(1/2) + 1 = 0}}}

Two ways to solve it. Pick whichever one
is the method you are studying now.

Method 1.

Change {{{x^(1/2)}}} to {{{sqrt(x)}}}

{{{2x - 3sqrt(x) + 1 = 0}}}

Isolate the radical term

{{{2x + 1 = 3sqrt(x)}}}

Square both sides of the equation:

{{{(2x + 1)^2 = (3sqrt(x))^2}}}

{{{(2x + 1)(2x + 1) = 9x }}}

{{{4x^2+2x+2x+1 = 9x }}}

{{{4x^2+4x+1 = 9x}}}

Get 0 on the right by subtracting 9x 
from both sides:

{{{4x^2-5x+1=0}}}

Factor the left side

{{{(4x-1)(x-1)=0}}}

Setting {{{4x-1=0}}} gives {{{x=1/4}}}

Setting {{{x-1=0}}} gives {{{x=1}}}

Both answers check.

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Method 2.

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

Let {{{x^(1/2) = w}}}

Then squaring both sides of that we have

   {{{x = w^2}}}

So the equation becomes

{{{2w^2 - 3w + 1 = 0}}}

Now we factor the left side:

{{{(2w-1)(w-1)=0}}}

{{{2w-1=0}}} gives {{{w=1/2}}}

{{{w-1=0}}} gives {{{x=1}}}

But we want x, not w, so since

{{{x = w^2}}}

For {{{w=1/2}}}

{{{x = w^2 = (1/2)^2 = 1/4}}}  

For {{{w=1}}}

{{{x = w^2 = 1^2 = 1}}}  

Edwin</pre>