How do you find the zeros, real and imaginary, of #y=x^2+32x+44# using the quadratic formula?

Question

Genesis Allen · Accepted Answer

Substitute the coefficients into the quadratic formula to find:

#x=-16+-2sqrt(53)#

#x^2+32x+44# is of the form #ax^2+bx+c# with #a=1#, #b=32# and #c=44#. This has zeros given by the quadratic formula: #x = (-b+-sqrt(b^2-4ac))/(2a)# #=(-32+-sqrt(32^2-(4xx1xx44)))/(2*1)# #=(-32+-sqrt(1024-176))/2# #=(-32+-sqrt(848))/2# #=(-32+-sqrt(4^2*53))/2# #=(-32+-4sqrt(53))/2# #=-16+-2sqrt(53)#

Savannah Anderson · Answer

undefined To find the zeros of the quadratic equation $ y = x^2 + 32x + 44 $ using the quadratic formula, first identify the coefficients: $ a = 1 $, $ b = 32 $, and $ c = 44 $. Then, plug these values into the quadratic formula:

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

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

$$ x = \frac{{-32 \pm \sqrt{{32^2 - 4 \cdot 1 \cdot 44}}}}{{2 \cdot 1}} $$

Simplify inside the square root:

$$ x = \frac{{-32 \pm \sqrt{{1024 - 176}}}}{2} $$
$$ x = \frac{{-32 \pm \sqrt{{848}}}}{2} $$

Now, calculate the square root:

$$ x = \frac{{-32 \pm \sqrt{{16 \cdot 53}}}}{2} $$
$$ x = \frac{{-32 \pm 4\sqrt{{53}}}}{2} $$

This gives us two solutions for $ x $:

$$ x_1 = \frac{{-32 + 4\sqrt{{53}}}}{2} $$
$$ x_2 = \frac{{-32 - 4\sqrt{{53}}}}{2} $$

Thus, the zeros of the quadratic equation $ y = x^2 + 32x + 44 $ are $ x_1 = \frac{{-32 + 4\sqrt{{53}}}}{2} $ and $ x_2 = \frac{{-32 - 4\sqrt{{53}}}}{2} $, which can also be simplified as $ x_1 = -16 + 2\sqrt{53} $ and $ x_2 = -16 - 2\sqrt{53} $.