Question

How do you find the zeros, if any, of `y= -4x^2 - 8`using the quadratic formula?

Answer 1

Since `y` is less than zero for all Real values of `x` there are no Real zeros.
However the quadratic formula can be used to determine the Complex zeros: `x=+-sqrt(2)i`

Explanation:

The quadratic formula tells us that for a quadratic in the form:
`color(white)("XXX")color(red)(a)x^2+color(blue)(b)+color(green)(c)=0`
the zeros occur at
`color(white)("XXX")x=(-color(blue)(b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)(a))`

Converting the given
`color(white)("XXX")y=-4x^2-8`
into the required form:
`color(white)("XXX")y=color(red)(""(-4))x^2+color(blue)(0)x+color(green)(""(-8))`

The zeros occur at
`color(white)("XXX")x=(color(blue)(0)+-sqrt(color(blue)(0)^2-4(color(red)(-4))(color(green)(-8))))/(2(color(red)(-4)))`

`color(white)("XxXX")=+-sqrt(-2)`

`color(white)("XxXX")=+-sqrt(2)i`

Answer 2

There is no 'Real Number' solution for `y=0` so the graph does not cross the x-axis.

However there is a solution for `y=0` within the set of numbers called Complex Numbers, and that is `x=+-i sqrt(2)`

The solution for `x=0` is `y=-8`

Explanation:

The x-intercepts occur at the points where the graph crosses the x-axis. This is when y=0.

Given:`" "y=-4x^2-8`

Substitute 0 for y giving:

`" "0=-4x^2-8`

Multiply both sides by `(-1)` giving

`" "0=4x^2+8`

Subtract 8 from both sides

`" "0-8=4x^2+8-8`

But `8-8=0`

`" "-8=4x^2`

Divide both sides by 4

`-8/4=4/4xx x^2`

But `4/4=1`

`-2=x^2`

Take the square root of each side

`x=+-sqrt(-2)`

There is no 'Real Number' solution to `x` so the graph does not cross the x-axis.

However there is a solution within the set of numbers called complex numbers and that is `x=+-i sqrt(2)`