Question

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

Answer 1

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`