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

Question

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

Kennedy Adams · Accepted 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# 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#

Isaiah Anthony · 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# 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)#

Nathan Axel · Answer

undefined To find the zeros of the quadratic equation $ y = -4x^2 - 8 $, you can use the quadratic formula, which is given as:

$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

In the equation $ y = -4x^2 - 8 $, $ a = -4 $, $ b = 0 $, and $ c = -8 $.

Substituting these values into the quadratic formula:

$$ x = \frac{{-0 \pm \sqrt{{0^2 - 4 \cdot (-4) \cdot (-8)}}}}{{2 \cdot (-4)}} $$

Simplify the expression under the square root:

$$ x = \frac{{\pm \sqrt{{-128}}}}{{-8}} $$

$$ x = \frac{{\pm \sqrt{{128}}}}{{-8}} $$

$$ x = \frac{{\pm 8\sqrt{2}}}{{-8}} $$

$$ x = \frac{{\pm \sqrt{2}}}{{-1}} $$

$$ x = \pm \sqrt{2} $$

Therefore, the zeros of the equation $ y = -4x^2 - 8 $ are $ x = -\sqrt{2} $ and $ x = \sqrt{2} $.

Avery Alexander · Answer

undefined To find the zeros of the quadratic equation $ y = -4x^2 - 8 $ using the quadratic formula, we first need to identify the coefficients $ a $, $ b $, and $ c $ in the general form $ ax^2 + bx + c $.

In this equation:
- $ a = -4 $
- $ b = 0 $ (since there's no term with $ x $)
- $ c = -8 $

Now, we can apply the quadratic formula:
$$ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $$

Substitute the values of $ a $, $ b $, and $ c $ into the formula:
$$ x = \frac{{-0 \pm \sqrt{{0^2 - 4(-4)(-8)}}}}{{2(-4)}} $$

Simplify:
$$ x = \frac{{\pm \sqrt{{-128}}}}{{-8}} $$

$$ x = \frac{{\pm \sqrt{{128}}}}{{8}} $$

$$ x = \frac{{\pm 8\sqrt{2}}}{{8}} $$

$$ x = \pm \sqrt{2} $$

So, the zeros of the quadratic equation $ y = -4x^2 - 8 $ are $ x = -\sqrt{2} $ and $ x = \sqrt{2} $.