How do you find the zeros, real and imaginary, of #y=-x^2-3x+11# using the quadratic formula?

1 Answer
Oct 18, 2017

Answer:

See a solution process below:

Explanation:

We can use the quadratic equation to solve this problem by equating the quadratic to #0# to find the roots or the zeros:

The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))#

Substituting:

#color(red)(-1)# for #color(red)(a)#

#color(blue)(-3)# for #color(blue)(b)#

#color(green)(11)# for #color(green)(c)# gives:

#x = (-color(blue)((-3)) +- sqrt(color(blue)((-3))^2 - (4 * color(red)(-1) * color(green)(11))))/(2 * color(red)(-1))#

#x = (color(blue)((-3)) +- sqrt(9 - (-44)))/(-2)#

#x = (color(blue)((-3)) +- sqrt(9 + 44))/(-2)#

#x = (-3 +- sqrt(53))/(-2)#

#x = (3 +- sqrt(53))/2#