Question 122284
Let {{{u=x^2}}} to get


{{{u^2-3u+2=0}}}



Let's use the quadratic formula to solve for u:



Starting with the general quadratic


{{{au^2+bu+c=0}}}


the general solution using the quadratic equation is:


{{{u = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{u^2-3*u+2=0}}} ( notice {{{a=1}}}, {{{b=-3}}}, and {{{c=2}}})





{{{u = (--3 +- sqrt( (-3)^2-4*1*2 ))/(2*1)}}} Plug in a=1, b=-3, and c=2




{{{u = (3 +- sqrt( (-3)^2-4*1*2 ))/(2*1)}}} Negate -3 to get 3




{{{u = (3 +- sqrt( 9-4*1*2 ))/(2*1)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{(-3)^2=-3*-3=9}}}.)




{{{u = (3 +- sqrt( 9+-8 ))/(2*1)}}} Multiply {{{-4*2*1}}} to get {{{-8}}}




{{{u = (3 +- sqrt( 1 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{u = (3 +- 1)/(2*1)}}} Simplify the square root (note: If you need help with simplifying the square root, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




{{{u = (3 +- 1)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{u = (3 + 1)/2}}} or {{{u = (3 - 1)/2}}}


Lets look at the first part:


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


{{{u=4/2}}} Add the terms in the numerator

{{{u=2}}} Divide


So one answer is

{{{u=2}}}




Now lets look at the second part:


{{{x=(3 - 1)/2}}}


{{{u=2/2}}} Subtract the terms in the numerator

{{{u=1}}} Divide


So another answer is

{{{u=1}}}


So our solutions are:

{{{u=2}}} or {{{u=1}}}


Now remember, we let {{{u=x^2}}}




So our solutions become




{{{x^2=2}}} or {{{x^2=1}}}




Now take the square root of both sides for each case



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




Answer:



So our answers are



{{{x=sqrt(2)}}}, {{{x=-sqrt(2)}}}, {{{x=1}}}, or {{{x=-1}}}