How do you find the roots, real and imaginary, of y= x^2 - x - (x-9)^2  using the quadratic formula?

Mar 19, 2017

We have real zero of the function, which is $\frac{81}{17}$

Explanation:

In the given function as we will see below terms relating to ${x}^{2}$ cancel out and hence, the function is a linear one and not quadratic and we do not need quadratic formula. Let us try to solve this.

$y = {x}^{2} - x - {\left(x - 9\right)}^{2}$

= ${x}^{2} - x - \left({x}^{2} - 18 x + 81\right)$ - using formula for ${\left(a - b\right)}^{2}$

= ${x}^{2} - x - {x}^{2} + 18 x - 81$

= ${\cancel{x}}^{2} - x - {\cancel{x}}^{2} + 18 x - 81$

= $17 x - 81$

Hence, we have real zero of the function which is $\frac{81}{17}$