Question 99573
{{{(x-1)/(x+2)=(2x-3)/(x+4)}}} Start with the given equation



{{{(x-1)(x+4)=(2x-3)(x+2)}}} Cross multiply



{{{x^2+3x-4=2x^2+x-6}}} Foil



{{{0=x^2-2x-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 {{{x^2-2*x-2=0}}} ( notice {{{a=1}}}, {{{b=-2}}}, and {{{c=-2}}})


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




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




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




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




{{{x = (2 +- sqrt( 12 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (2 +- 2*sqrt(3))/(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 = (2 +- 2*sqrt(3))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (2 + 2*sqrt(3))/2}}} or {{{x = (2 - 2*sqrt(3))/2}}}



Now break up the fraction



{{{x=+2/2+2*sqrt(3)/2}}} or {{{x=+2/2-2*sqrt(3)/2}}}



Simplify



{{{x=1+sqrt(3)}}} or {{{x=1-sqrt(3)}}}



So these expressions approximate to


{{{x=2.73205080756888}}} or {{{x=-0.732050807568877}}}



So our solutions are:

{{{x=2.73205080756888}}} or {{{x=-0.732050807568877}}}


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


{{{ graph( 500, 500, -10.7320508075689, 12.7320508075689, -10.7320508075689, 12.7320508075689,1*x^2+-2*x+-2) }}}


when we use the root finder feature on a calculator, we find that {{{x=2.73205080756888}}} and {{{x=-0.732050807568877}}}.So this verifies our answer