Question

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

Answer 1

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

Explanation:

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

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

Answer 2

`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`