How do you find the zeros, real and imaginary, of #y=x^2-31x-9# using the quadratic formula?

Question

Abigail Allen · Accepted Answer

#x_1=(+31+sqrt((961+36)))/(2)=31.2877#
and #x_2=(+31-sqrt((961+36)))/(2)=-0.287653# from the given: #y=x^2-31x-9# #a=1#, #b=-31# , #c=-9# #x=(-b+-sqrt(b^2-4ac))/(2a)# #x=(--31+-sqrt((-31)^2-4(1)(-9)))/(2*1)# #x=(+31+-sqrt((961+36)))/(2)# #x_1=(+31+sqrt((961+36)))/(2)=31.2877# and #x_2=(+31-sqrt((961+36)))/(2)=-0.287653# have a nice day from the Philippines!

Savannah Anderson · Answer

undefined To find the zeros of the quadratic equation $y = x^2 - 31x - 9$, you can use the quadratic formula, which is given by $x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$, where $a$, $b$, and $c$ are the coefficients of the quadratic equation $ax^2 + bx + c$.

For the given equation $y = x^2 - 31x - 9$:
$a = 1$, $b = -31$, and $c = -9$.

Plugging these values into the quadratic formula:
$x = \frac{{-(-31) \pm \sqrt{{(-31)^2 - 4(1)(-9)}}}}{{2(1)}}$.

Solving this equation:
$x = \frac{{31 \pm \sqrt{{961 + 36}}}}{2}$,
$x = \frac{{31 \pm \sqrt{{997}}}}{2}$.

Thus, the zeros of the equation are:
$x = \frac{{31 + \sqrt{{997}}}}{2}$ (real) and $x = \frac{{31 - \sqrt{{997}}}}{2}$ (real).