How do you find the roots, real and imaginary, of #y= x^2 - 121 # using the quadratic formula?

1 Answer
Jun 8, 2018

Answer:

The two solutions are #11# and #-11#

Explanation:

Compare your equation, #x^2-121#, with the generic equation #ax^2+bx+c#. Your equation is obtained by choosing #a=1#, #b=0#, #c=-121#.

The quadratic formula states that the solution of a quadratic equation are

#x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}#

Plug your coefficients:

#x_{1,2} = \frac{0\pm\sqrt{0^2-4*1*(-121)}}{2*1} = \pm\frac{sqrt{484}}{2} = \pm\frac{22}{2} = \pm11#

So, the two solutions are #11# and #-11#