# How do you find the roots, real and imaginary, of y=8x^2+7x-(3x-1)^2  using the quadratic formula?

Apr 10, 2016

Zeros of function are $x = \frac{13}{2} - \frac{\sqrt{165}}{2}$ and $x = \frac{13}{2} + \frac{\sqrt{165}}{2}$

#### Explanation:

The roots of general form of equation $a {x}^{2} + b x + c = 0$ are zeros of the general form of equation $y = a {x}^{2} + b x + c$.

Simplifying the given equation $y = 8 {x}^{2} + 7 x - {\left(3 x - 1\right)}^{2}$, gives us

$y = 8 {x}^{2} + 7 x - \left(9 {x}^{2} - 6 x + 1\right) = - {x}^{2} + 13 x - 1$

we see that $a = - 1$, $b = 13$ and $c = - 1$.

As discriminant ${b}^{2} - 4 a c = {\left(13\right)}^{2} - 4 \left(- 1\right) \left(- 1\right) = 169 - 4 = 165$

As discriminant is positive but not a complete square, we have real but irrational roots.

Hence, using quadratic formula $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$,

the zeros are the given equation are $x = \frac{- \left(13\right) \pm \sqrt{165}}{2 \cdot \left(- 1\right)}$

or Zeros of function are $x = \frac{13}{2} - \frac{\sqrt{165}}{2}$ and $x = \frac{13}{2} + \frac{\sqrt{165}}{2}$