How do you find the zeros, real and imaginary, of #y= -9x^2-28x-3# using the quadratic formula?

1 Answer
Jul 26, 2018

Answer:

The zeros are #x = -3, -1/9#.

Explanation:

#y = -9x^2 - 28x - 3#

This equation is in standard quadratic form, or #y = color(red)(a)x^2 + color(blue)(b)x + color(magenta)(c)#, where #color(red)(a = -9)#, #color(blue)(b = -28)#, and #color(magenta)(c = -3)#.

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

Plug in the values and solve:
#x = (-(color(blue)(-28)) +- sqrt((color(blue)(-28))^2 - 4(color(red)(-9))(color(magenta)(-3))))/(2(color(red)(-9)))#:

#x = (28 +- sqrt(784 - 108))/(-18)#

#x = (28 +- sqrt(676))/(-18)#

#x = (28 +- 26)/(-18)#

#x = (28+26)/(-18)#, #(28-26)/(-18)#

#x = 54/-18#, #2/-18#

#x = -3, -1/9#

Hope this helps!