How do you find the x intercepts for #y = -2x^2 + 5x - 1#?

2 Answers
Jun 30, 2015

Solve #y = -2x^2 + 5x - 1 = 0#

Explanation:

#D = d^2 = 25 - 8 = 17 -> d = +- sqrt17#

#x = 5/4 +- (sqrt17)/4#

Jul 1, 2015

#x = 5/4 +-sqrt(17)/4#

Explanation:

This answer is the same as given by Nghi N. I am simply providing a more detailed explanation.

Since the x-intercepts occur when #y=0# we solve the given equation with #0# in place of #y# to find the x-intercepts.

#−2x^2+5x−1=0#

Any parabolic equation of the form #ax^2+bx+c=0#
can be solved using the quadratic formula:
#color(white)("XXXX")##x = (-b+-sqrt(b^2-4ac))/(2a)#

Applying the values from our example:
#color(white)("XXXX")##x = (-5+-sqrt(5^2-4(-2)(-1)))/(2(-2))#

#color(white)("XXXX")##x = (-5 +- sqrt(25-8))/(-4)#

#color(white)("XXXX")##x = 5/4 +-sqrt(17)/4#