Question 88263
{{{3x^2 = 11x + 4}}}


{{{3x^2 - 11x - 4 =0 }}} Get everything to one side


Now let's use the quadratic formula to solve for x:



Starting with the general quadratic


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


the general form of the quadratic equation is:


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


So lets solve {{{3*x^2-11*x-4=0}}} (notice {{{a=3}}}, {{{b=-11}}}, and {{{c=-4}}})


{{{x = (--11 +- sqrt( (-11)^2-4*3*-4 ))/(2*3)}}} Plug in a=3, b=-11, and c=-4




{{{x = (11 +- sqrt( (-11)^2-4*3*-4 ))/(2*3)}}} Negate -11 to get 11




{{{x = (11 +- sqrt( 121-4*3*-4 ))/(2*3)}}} Square -11 to get 121




{{{x = (11 +- sqrt( 121+48 ))/(2*3)}}} Multiply {{{-4*-4*3}}} to get {{{48}}}




{{{x = (11 +- sqrt( 169 ))/(2*3)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (11 +- 13)/(2*3)}}} Simplify the square root




{{{x = (11 +- 13)/6}}} Multiply 2 and 3 to get 6


So now the expression breaks down into two parts


{{{x = (11 + 13)/6}}} or {{{x = (11 - 13)/6}}}


Lets look at the first part:


{{{x=24/6}}} Add the terms in the numerator

{{{x=4}}} Divide


So one answer is

{{{x=4}}}

Now lets look at the second part:


{{{x=-2/6}}} Subtract the terms in the numerator

{{{x=-1/3}}} Divide


So another answer is

{{{x=-1/3}}}


So our solutions are:

{{{x=4}}} or {{{x=-1/3}}}


Notice when we graph {{{3*x^2-11*x-4}}} we get:


{{{ graph( 500, 500, -11, 14, -11, 14,3*x^2+-11*x+-4) }}}


and we can see that the roots are {{{x=4}}} and {{{x=-1/3}}}. This verifies our answer