Question 92553
{{{4x^2 = 13x + 12}}}


{{{4x^2 -13x - 12=0}}} 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 {{{4*x^2-13*x-12=0}}} ( notice {{{a=4}}}, {{{b=-13}}}, and {{{c=-12}}})


{{{x = (--13 +- sqrt( (-13)^2-4*4*-12 ))/(2*4)}}} Plug in a=4, b=-13, and c=-12




{{{x = (13 +- sqrt( (-13)^2-4*4*-12 ))/(2*4)}}} Negate -13 to get 13




{{{x = (13 +- sqrt( 169-4*4*-12 ))/(2*4)}}} Square -13 to get 169  (note: remember when you square -13, you must square the negative as well. This is because {{{(-13)^2=-13*-13=169}}}.)




{{{x = (13 +- sqrt( 169+192 ))/(2*4)}}} Multiply {{{-4*-12*4}}} to get {{{192}}}




{{{x = (13 +- sqrt( 361 ))/(2*4)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (13 +- 19)/(2*4)}}} 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 = (13 +- 19)/8}}} Multiply 2 and 4 to get 8


So now the expression breaks down into two parts


{{{x = (13 + 19)/8}}} or {{{x = (13 - 19)/8}}}


Lets look at the first part:


{{{x=(13 + 19)/8}}}


{{{x=32/8}}} Add the terms in the numerator

{{{x=4}}} Divide


So one answer is

{{{x=4}}}




Now lets look at the second part:


{{{x=(13 - 19)/8}}}


{{{x=-6/8}}} Subtract the terms in the numerator

{{{x=-3/4}}} Divide


So another answer is

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


So our solutions are:

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


Notice when we graph {{{4*x^2-13*x-12}}}, we get:


{{{ graph( 500, 500, -13, 14, -13, 14,4*x^2+-13*x+-12) }}}


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