Question 117560
{{{x^2+2x-1=0}}} To find the x-intercept, we must set y equal to zero and solve for x



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





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




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




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




{{{x = (-2 +- sqrt( 8 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




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


So now the expression breaks down into two parts


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



Now break up the fraction



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



Simplify



{{{x=-1+sqrt(2)}}} or {{{x=-1-sqrt(2)}}}



So these expressions approximate to


{{{x=0.414213562373095}}} or {{{x=-2.41421356237309}}}



So our solutions are:

{{{x=0.414213562373095}}} or {{{x=-2.41421356237309}}}


Notice when we graph {{{x^2+2*x-1}}}, we get:


{{{ graph( 500, 500, -12.4142135623731, 10.4142135623731, -12.4142135623731, 10.4142135623731,1*x^2+2*x+-1) }}}


when we use the root finder feature on a calculator, we find that {{{x=0.414213562373095}}} and {{{x=-2.41421356237309}}}.So this verifies our answer