Question 118394
{{{3(x+2) + 4(x-1) = 7(x+2)(x-1)}}} Start with the given equation



{{{3(x+2) + 4(x-1) = 7(x^2+x-2)}}} Foil



{{{3x+6 + 4x-4 = 7x^2+7x-14}}} Distribute




{{{3x+6 + 4x-4 - 7x^2-7x+14=0}}} Get all terms to the left side



{{{-7x^2+16=0}}} Combine like terms




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 {{{-7*x^2+16=0}}} (note: since the polynomial does not have an "x" term, the 2nd coefficient is zero. In other words, b=0. So that means the polynomial really looks like {{{-7*x^2+0*x+16=0}}}  notice {{{a=-7}}}, {{{b=0}}}, and {{{c=16}}})





{{{x = (0 +- sqrt( (0)^2-4*-7*16 ))/(2*-7)}}} Plug in a=-7, b=0, and c=16




{{{x = (0 +- sqrt( 0-4*-7*16 ))/(2*-7)}}} Square 0 to get 0  




{{{x = (0 +- sqrt( 0+448 ))/(2*-7)}}} Multiply {{{-4*16*-7}}} to get {{{448}}}




{{{x = (0 +- sqrt( 448 ))/(2*-7)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-0 +- 8*sqrt(7))/(2*-7)}}} 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 = (-0 +- 8*sqrt(7))/-14}}} Multiply 2 and -7 to get -14


So now the expression breaks down into two parts


{{{x = (-0 + 8*sqrt(7))/-14}}} or {{{x = (-0 - 8*sqrt(7))/-14}}}



Now break up the fraction



{{{x=-0/-14+8*sqrt(7)/-14}}} or {{{x=-0/-14-8*sqrt(7)/-14}}}



Simplify



{{{x=0-4*sqrt(7)/7}}} or {{{x=0+4*sqrt(7)/7}}}



So these expressions approximate to


{{{x=-1.51185789203691}}} or {{{x=1.51185789203691}}}



So our solutions are:

{{{x=-1.51185789203691}}} or {{{x=1.51185789203691}}}


Notice when we graph {{{-7*x^2+16}}}, we get:


{{{ graph( 500, 500, -11.5118578920369, 11.5118578920369, -11.5118578920369, 11.5118578920369,-7*x^2+0*x+16) }}}


when we use the root finder feature on a calculator, we find that {{{x=-1.51185789203691}}} and {{{x=1.51185789203691}}}.So this verifies our answer