How do you find the zeros, real and imaginary, of #y= -x^2-12x-8 # using the quadratic formula?

Question

Isaiah Anthony · Accepted Answer

#x=-6+2sqrt7;# or #x=-6-2sqrt7#

When using the quadratic formula, the zeros are the two values for #x#. When solving quadratic equations, the equation must be equal to #0#.

Your equation: #y=-x^2-12x-8#

Set #y# equal to zero.

#0=-x^2-12x-8#

Multiply by #-1# to make #a# positive.

#x^2+12x+-8#

#-x^2-12x-8# is a quadratic equation in the form of #"color(red)(a)^2+color(blue)bx+color(green)(c)#, where the coefficients are #"color(red)(a)=(1)#, #color(blue)(b)=12#, #color(green)(c)=8#.

Quadratic equation

#x=(color(blue)(-b)+-sqrt(color(blue)(b)^2-4color(red)(a)color(green)(c)))/(2color(red)a)#

Substitute the known values into the equation.

#x=("-(color(blue)(12))+-sqrt(color(blue)(12)^2-4(color(red)(1))(color(green)(8))))/((2*color(red)(1)))#

#x=(-12+-sqrt(144-(32)))/(-2)#

#=(-12+-sqrt(112))/(-2)#

Factor #sqrt112#.

#sqrt112=sqrt(16xx7)=4sqrt7#

#x=(-12+-4sqrt7)/2#

Simplifyby factoring out #2# in #-12, +-4, and 2#.

#x=-6+-2sqrt7#

#x=-6+2sqrt7;# or #x=-6-2sqrt7#

Ethan Adkins · Answer

undefined To find the zeros of the quadratic equation $y = -x^2 - 12x - 8$ using the quadratic formula:

1. Identify the coefficients $a$, $b$, and $c$. In this equation, $a = -1$, $b = -12$, and $c = -8$.

2. Substitute the values of $a$, $b$, and $c$ into the quadratic formula:

$x = \frac{{-(-12) \pm \sqrt{{(-12)^2 - 4(-1)(-8)}}}}{{2(-1)}}$

3. Simplify the expression inside the square root:

$x = \frac{{12 \pm \sqrt{{144 - 32}}}}{-2}$

$x = \frac{{12 \pm \sqrt{{112}}}}{-2}$

4. Now, simplify the expression under the square root, if possible:

$x = \frac{{12 \pm \sqrt{{16 \times 7}}}}{-2}$

$x = \frac{{12 \pm 4\sqrt{{7}}}}{-2}$

5. Divide both the numerator and denominator by -2:

$x = \frac{{-6 \pm 2\sqrt{7}}}{1}$

Therefore, the zeros of the quadratic equation $y = -x^2 - 12x - 8$ are $x = -6 + 2\sqrt{7}$ and $x = -6 - 2\sqrt{7}$. These are real numbers. There are no imaginary zeros.