Question 123070
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 {{{3*x^2+4*x-9=0}}} ( notice {{{a=3}}}, {{{b=4}}}, and {{{c=-9}}})





{{{x = (-4 +- sqrt( (4)^2-4*3*-9 ))/(2*3)}}} Plug in a=3, b=4, and c=-9




{{{x = (-4 +- sqrt( 16-4*3*-9 ))/(2*3)}}} Square 4 to get 16  




{{{x = (-4 +- sqrt( 16+108 ))/(2*3)}}} Multiply {{{-4*-9*3}}} to get {{{108}}}




{{{x = (-4 +- sqrt( 124 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-4 +- 2*sqrt(31))/(2*3)}}} 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 = (-4 +- 2*sqrt(31))/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (-4 + 2*sqrt(31))/6}}} or {{{x = (-4 - 2*sqrt(31))/6}}}



Now break up the fraction



{{{x=-4/6+2*sqrt(31)/6}}} or {{{x=-4/6-2*sqrt(31)/6}}}



Simplify



{{{x=-2 / 3+sqrt(31)/3}}} or {{{x=-2 / 3-sqrt(31)/3}}}



So these expressions approximate to


{{{x=1.18925478761001}}} or {{{x=-2.52258812094334}}}



So our solutions are:

{{{x=1.18925478761001}}} or {{{x=-2.52258812094334}}}


Notice when we graph {{{3*x^2+4*x-9}}}, we get:


{{{ graph( 500, 500, -12.5225881209433, 11.18925478761, -12.5225881209433, 11.18925478761,3*x^2+4*x+-9) }}}


when we use the root finder feature on a calculator, we find that {{{x=1.18925478761001}}} and {{{x=-2.52258812094334}}}.So this verifies our answer