Question 110028: Find the x-intercepts.
y=x^2+5x+2
Please if you would take me through the steps? Thank you. Found 2 solutions by stanbon, edjones:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! Find the x-intercepts.
y=x^2+5x+2
-------------
Let y = 0 and solve for "x".
x^2+5x+2 = 0
x = [-5 +- sqrt(25-4*2)]/2
x = [-5 +- sqrt(17)]/2
x = (-5 + sqrt(17))/2 or x = (-5 - sqrt(17))/2
========================
=========================
Cheers,
Stan H.
You can put this solution on YOUR website! the x intercepts occur when y=0
x^2+5x+2=0
x^2+5x =-2
x^2+5x+25/4=-2+ 25/4 Complete the square by adding 25/4 to both sides.
(x+ 5/2)^2=17/4
x+5/2=+-1/2sqrt(17)
x=-5/2+-1/2sqrt(17)
x= -.438447, x= -4.56155
Ed
(