# How do you find the roots, real and imaginary, of y=-5x^2 +(x -4)^2  using the quadratic formula?

May 26, 2016

We have two real zeros, which are irrational. These are $- 1 \pm \sqrt{5}$

#### Explanation:

Quadratic formula uses general form of quadratic equation $a {x}^{2} + b x + c = 0$ abd gives the solution of this equation as $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. If $\alpha$ and$\beta$ are roots of this equation, then they are also zeros of the function $y = a {x}^{2} + b x + c$.

Hence, we should first convert the function $y = - 5 {x}^{2} + {\left(x - 4\right)}^{2}$, to general form. Simplifying it we get,

$y = - 5 {x}^{2} + {x}^{2} - 8 x + 16$

or $y = - 4 {x}^{2} - 8 x + 16$ and hence zeros of this function are

$x = \frac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \cdot \left(- 4\right) \cdot 16}}{2 \cdot \left(- 4\right)}$

or $x = \frac{8 \pm \sqrt{64 + 256}}{- 8} = \frac{8 \pm \sqrt{320}}{- 8}$

or $x = \frac{8 \pm 8 \sqrt{5}}{- 8} = - 1 \pm \sqrt{5}$

Hence, we have two real zeros, which are irrational.