How do you find the zeros, real and imaginary, of y=-x^2-6x+11 using the quadratic formula?

1 Answer
May 8, 2018

(-3 + 2sqrt5, 0) and (-3 - 2sqrt5, 0)

Explanation:

y = -x^2 - 6x + 11

To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c) = 0, which this equation is.

So we know that:
color(red)(a = -1)

color(magenta)(b = -6)

color(blue)(c = 11)

The quadratic formula is x = (-color(magenta)(b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a)).

Now we can plug in the values for color(red)(a), color(magenta)(b), and color(blue)(c) into the quadratic formula:

x = (-(color(magenta)(-6)) +- sqrt((color(magenta)(-6))^2 - 4(color(red)(-1))(color(blue)(11))))/(2(color(red)(-1)))

Simplify:
x = (6 +- sqrt(36 + 44))/-2

x = (6 +- sqrt(80))/-2

x = (6 +- 4sqrt5)/-2

x = -3 +- 2sqrt5

This is the same thing as:
x = -3 + 2sqrt5 and x = -3 - 2sqrt5
because +- means "plus or minus."

This means the zeros are at:
(-3 + 2sqrt5, 0) and (-3 - 2sqrt5, 0)

Hope this helps!