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

Question

1966 views around the world

You can reuse this answer
Creative Commons License

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]}

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