Question 217614
{{{ x^(1/2)-3x^(1/4)+2=0 }}}
There are a couple of keys to a relatively simple solution to this problem:<ul><li>Recognize that 1/2 is two times 1/4</li><li>Recognize that because 1/2 is 2 times 1/4 and because of the rules for exponents, {{{x^(1/2) = (x^(1/4))^2}}}</li></ul>
Once you get that {{{x^(1/2) = (x^(1/4))^2}}}, the rest of the problem is fairly simple. Without seeing this, the problem would be very difficult.<br>
Your equation is of the form {{{q^2 - 3q +2 = 0}}} (with {{{q = x^(1/4)}}}). And we solve this by factoring (or with the quadratic formula):
{{{(q-2)(q-1)=0}}}
The only way for this product (multiplication) to be zero is if one of the factors is zero. So we simply state this with equations:
q-2 = 0 or q-1 = 0
Solving each we get:
q = 2 or q = 1
Of course we're not interested in "q". So we substitute back in for q:
{{{x^(1/4) = 2}}} or {{{x^(1/4) = 1}}}
Raising each side of each equation to the 4th power:
x = 16 or x = 1<br>
Once you've done a few of these you will not need a substitute variable. You'll see that {{{ x^(1/2)-3x^(1/4)+2=0 }}} will factor into {{{(x^(1/4)-2)(x^(1/4)-1) = 0}}}. And once you have a product equal to zero you are pretty close to the final solution.