How do you find the roots, real and imaginary, of #y= -x^2 - 12x + 4 # using the quadratic formula?

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Answer 1 · 2017-10-02T12:13:08

See a solution process below:

To find the roots of the equation set #y# equal to #0# and solve for #x#.

#-x^2 - 12x + 4 = 0#

Now, we can use the quadratic equation to solve this problem:

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)(-12)# for #color(blue)(b)#

#color(green)(4)# for #color(green)(c)# gives:

#x = (-color(blue)((-12)) +- sqrt(color(blue)((-12))^2 - (4 * color(red)(-1) * color(green)(4))))/(2 * color(red)(-1))#

#x = (12 +- sqrt(144 - (-16)))/(-2)#

#x = (12)/-2 +- (sqrt(144 + 16))/(-2)#

#x = -6 +- (sqrt(160))/(-2)#

#x = -6 - (sqrt(16 * 10))/(-2)# and #x = -6 + (sqrt(16 * 10))/(-2)#

#x = -6 - (sqrt(16)sqrt(10))/(-2)# and #x = -6 + (sqrt(16)sqrt(10))/(-2)#

#x = -6 - (4sqrt(10))/(-2)# and #x = -6 + (4sqrt(10))/(-2)#

#x = -6 - (-2sqrt(10))# and #x = -6 + (-2sqrt(10))#

#x = -6 + 2sqrt(10)# and #x = -6 - 2sqrt(10)#