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

2 Answers
May 13, 2016

Answer:

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#

May 13, 2016

Answer:

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)#