How do you find the zeros of # y = -x^2 - 1/11x + 1/7 # using the quadratic formula?

Question

How do you find the zeros of # y =  -x^2 - 1/11x + 1/7 # using the quadratic formula?

Ethan Adkins · Accepted Answer

Zeros are #7/22-7/2sqrt(491/847)# and #7/22+7/2sqrt(491/847)# Quadratic formula gives the zeros of #y=ax^2+bx+c# as #x=(-b+-sqrt(b^2-4ac))/(2a)#. Hence zeros of #y=-x^2-1/11x+1/7# are #x=(-(-1/11)+-sqrt((-1/11)^2-4(-1)(1/7)))/(2(1/7))# = #((1/11)+-sqrt(1/121+4/7))/(2/7)# = #(1/11+-sqrt((7+484)/847))xx7/2# = #7/22+-7/2sqrt(491/847)# Hence zeros are #7/22-7/2sqrt(491/847)# and #7/22+7/2sqrt(491/847)#

Avery Alexander · Answer

undefined To find the zeros of $ y = -x^2 - \frac{1}{11}x + \frac{1}{7} $ using the quadratic formula, first identify the coefficients:
$ a = -1 $,
$ b = -\frac{1}{11} $, and
$ c = \frac{1}{7} $.

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

Substitute the coefficients:
$$ x = \frac{{-\left(-\frac{1}{11}\right) \pm \sqrt{{\left(-\frac{1}{11}\right)^2 - 4(-1)\left(\frac{1}{7}\right)}}}}{{2(-1)}} $$

Simplify:
$$ x = \frac{{\frac{1}{11} \pm \sqrt{{\frac{1}{121} + \frac{4}{7}}}}}{{-2}} $$

$$ x = \frac{{\frac{1}{11} \pm \sqrt{{\frac{1}{121} + \frac{28}{121}}}}}{{-2}} $$

$$ x = \frac{{\frac{1}{11} \pm \sqrt{{\frac{29}{121}}}}}{{-2}} $$

$$ x = \frac{{\frac{1}{11} \pm \frac{\sqrt{29}}{11}}}{{-2}} $$

$$ x = \frac{1 \pm \sqrt{29}}{-22} $$

So, the zeros are $ x = \frac{1 + \sqrt{29}}{-22} $ and $ x = \frac{1 - \sqrt{29}}{-22} $.