Question 91979
First let y=0


{{{x^2+x-6=0}}}


Now let's use the quadratic formula to find the x-intercepts:



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+x-6=0}}} ( notice {{{a=1}}}, {{{b=1}}}, and {{{c=-6}}})


{{{x = (-1 +- sqrt( (1)^2-4*1*-6 ))/(2*1)}}} Plug in a=1, b=1, and c=-6




{{{x = (-1 +- sqrt( 1-4*1*-6 ))/(2*1)}}} Square 1 to get 1  




{{{x = (-1 +- sqrt( 1+24 ))/(2*1)}}} Multiply {{{-4*-6*1}}} to get {{{24}}}




{{{x = (-1 +- sqrt( 25 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{x = (-1 +- 5)/(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 = (-1 +- 5)/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-1 + 5)/2}}} or {{{x = (-1 - 5)/2}}}


Lets look at the first part:


{{{x=(-1 + 5)/2}}}


{{{x=4/2}}} Add the terms in the numerator

{{{x=2}}} Divide


So one answer is

{{{x=2}}}




Now lets look at the second part:


{{{x=(-1 - 5)/2}}}


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

{{{x=-3}}} Divide


So another answer is

{{{x=-3}}}


So our solutions are:

{{{x=2}}} or {{{x=-3}}}


Notice when we graph {{{x^2+x-6}}}, we get:


{{{ graph( 500, 500, -13, 12, -13, 12,1*x^2+1*x+-6) }}}


and we can see that the roots are {{{x=2}}} and {{{x=-3}}}. This verifies our answer