How do you find the roots, real and imaginary, of y= (45-x)x  using the quadratic formula?

May 7, 2016

$x = 0$, $x = 45$

Explanation:

Without using the quadratic formula, simply equate the equation to $0$

$\left(45 - x\right) x = 0$

For a product to be $0$, either factor should be 0

$\implies 45 - x = 0$
$\implies x = 45$

$\implies x = 0$

If you really need to get the zeroes using the quadratic formula.

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$y = \left(45 - x\right) x$

$\implies y = - {x}^{2} + 45 x + 0$

$\implies a = - 1$
$\implies b = 45$
$\implies c = 0$

$x = \left(- 45 \pm \sqrt{{45}^{2} - 4 \left(- 1\right) \left(0\right)}\right) \left(2 \left(- 1\right)\right)$

$\implies x = \frac{- 45 \pm \sqrt{{45}^{2} - 0}}{-} 2$

$\implies x = \frac{- 45 \pm 45}{-} 2$

$\implies x = \frac{- 45 + 45}{-} 2$
$\implies x = \frac{0}{-} 2 = 0$

$\implies x = \frac{- 45 - 45}{-} 2$
$\implies x = - \frac{90}{-} 2$
$\implies x = 45$