Question 102538
To find the x-intercepts, let y=0 and solve for x


{{{0=x^2+5x+2}}}


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 {{{x^2+5*x+2=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=2}}})





{{{x = (-5 +- sqrt( (5)^2-4*1*2 ))/(2*1)}}} Plug in a=1, b=5, and c=2




{{{x = (-5 +- sqrt( 25-4*1*2 ))/(2*1)}}} Square 5 to get 25  




{{{x = (-5 +- sqrt( 25+-8 ))/(2*1)}}} Multiply {{{-4*2*1}}} to get {{{-8}}}




{{{x = (-5 +- sqrt( 17 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)





{{{x = (-5 +- sqrt(17))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{x = (-5 + sqrt(17))/2}}} or {{{x = (-5 - sqrt(17))/2}}}



Now break up the fraction



{{{x=-5/2+sqrt(17)/2}}} or {{{x=-5/2-sqrt(17)/2}}}




So these expressions approximate to


{{{x=-0.43844718719117}}} or {{{x=-4.56155281280883}}}



So our solutions are:

{{{x=-0.43844718719117}}} or {{{x=-4.56155281280883}}}


Notice when we graph {{{x^2+5*x+2}}}, we get:


{{{ graph( 500, 500, -14.5615528128088, 9.56155281280883, -14.5615528128088, 9.56155281280883,1*x^2+5*x+2) }}}


when we use the root finder feature on a calculator, we find that {{{x=-0.43844718719117}}} and {{{x=-4.56155281280883}}}.So this verifies our answer