Question 99578
{{{2/x+2/(x-1)=1}}} Start with the given equation



{{{x(x-1)(2/x+2/(x-1))=x(x-1)(1)}}} Multiply both sides by the LCD {{{x(x+1)}}}



{{{2(x-1)+2x=x(x-1)}}} Distribute and multiply 



{{{2x-2+2x=x^2-x}}} Distribute again



{{{0=x^2-5x+2}}} Get all terms to one side



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



Starting with the general quadratic


{{{ax^2+bx+c=0}}}


the general solution using the quadratic equation is:


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


So lets solve {{{0=x^2-5*x+2}}} ( notice {{{a=1}}}, {{{b=-5}}}, and {{{c=2}}})


{{{x = (--5 +- sqrt( (-5)^2-4*1*2 ))/(2*1)}}} Plug in a=1, b=-5, and c=2




{{{x = (5 +- sqrt( (-5)^2-4*1*2 ))/(2*1)}}} Negate -5 to get 5




{{{x = (5 +- sqrt( 25-4*1*2 ))/(2*1)}}} Square -5 to get 25  (note: remember when you square -5, you must square the negative as well. This is because {{{(-5)^2=-5*-5=25}}}.)




{{{x = (5 +- sqrt( 25+-8 ))/(2*1)}}} Multiply {{{-4*2*1}}} to get {{{-8}}}




{{{x = (5 +- sqrt( 17 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (5 +- sqrt(17))/(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>)




{{{x = (5 +- sqrt(17))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (5 + sqrt(17))/2}}} or {{{x = (5 - sqrt(17))/2}}}



Now break up the fraction



{{{x=+5/2+sqrt(17)/2}}} or {{{x=+5/2-sqrt(17)/2}}}



Simplify



{{{x=5 / 2+sqrt(17)/2}}} or {{{x=5 / 2-sqrt(17)/2}}}



So these expressions approximate to


{{{x=4.56155281280883}}} or {{{x=0.43844718719117}}}



So our solutions are:

{{{x=4.56155281280883}}} or {{{x=0.43844718719117}}}


Notice when we graph {{{x^2-5*x+2}}}, we get:


{{{ graph( 500, 500, -9.56155281280883, 14.5615528128088, -9.56155281280883, 14.5615528128088,1*x^2+-5*x+2) }}}


when we use the root finder feature on a calculator, we find that {{{x=4.56155281280883}}} and {{{x=0.43844718719117}}}.So this verifies our answer