How do you find the zeros, if any, of #y= -8x^2 -x-2#using the quadratic formula?

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Answer 1 · 2018-05-24T12:25:14

See a solution process below:

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)(-8)# for #color(red)(a)#

#color(blue)(-1)# for #color(blue)(b)#

#color(green)(-2)# for #color(green)(c)# gives:

#x = (-color(blue)(-1) +- sqrt(color(blue)(-1)^2 - (4 * color(red)(-8) * color(green)(-2))))/(2 * color(red)(-8))#

#x = (1 +- sqrt(1 - (64)))/-16#

#x = (1 +- sqrt(-63))/-16#

Because the square root of a negative number is not a Real Number there are no zeros for this equation.

Graphing this equation shows:

graph{(y+8x^2+x+2)=0 [-20, 20, -15, 5]}