How do you find the zeros, real and imaginary, of #y=x^2+32x-4# using the quadratic formula?

Question

Julia Adams · Accepted Answer

#x = -16 +- 2sqrt65#

#y = x^2 + 32x - 4#

To use the quadratic formula to find the zeroes, we need to make sure the equation is written in the form #color(red)(a)x^2 + color(magenta)(b)x + color(blue)(c) = 0#, which this equation is.

So we know that: #color(red)(a = 1)#

#color(magenta)(b = 32)#

#color(blue)(c = -4)#

The quadratic formula is #x = (-color(magenta)(b) +- sqrt(color(magenta)(b)^2 - 4color(red)(a)color(blue)(c)))/(2color(red)(a))#.

Now we can plug in the values for #color(red)(a)#, #color(magenta)(b)#, and #color(blue)(c)# into the quadratic formula:

#x = (-color(magenta)(32) +- sqrt((color(magenta)(32))^2 - 4(color(red)(1))(color(blue)(-4))))/(2(color(red)(1)))#

Simplify: #x = (-32 +- sqrt(1024 + 16))/2#

#x = (-32 +- sqrt(1040))/2#

#x = (-32 +- 4sqrt65)/2#

#x = -16 +- 2sqrt65#

This is the same thing as: #x = -16 + 2sqrt65# and #x = -16 - 2sqrt65# because #+-# means "plus or minus."

Hope this helps!

Eva Adamson · Answer

undefined To find the zeros of the quadratic equation $y = x^2 + 32x - 4$ using the quadratic formula, we use the formula:

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

Where:
- $a = 1$
- $b = 32$
- $c = -4$

Plugging these values into the quadratic formula:

$$x = \frac{{-32 \pm \sqrt{{32^2 - 4 \cdot 1 \cdot (-4)}}}}{{2 \cdot 1}}$$

Solving under the square root:

$$32^2 - 4 \cdot 1 \cdot (-4) = 1024$$

$$x = \frac{{-32 \pm \sqrt{{1024}}}}{{2}}$$

$$x = \frac{{-32 \pm 32}}{{2}}$$

This gives two solutions:

$$x_1 = \frac{{-32 + 32}}{{2}} = 0$$
$$x_2 = \frac{{-32 - 32}}{{2}} = -32$$

So, the zeros of the equation are $x = 0$ and $x = -32$. Both are real zeros.