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

Question

1642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-07T03:23:29

$x = 0$ , $x = 45$

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

$(45 - x)x = 0$

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

$=> 45 - x = 0$
$=> x = 45$

$=> x = 0$

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

$x = (-b +- sqrt(b^2 - 4ac))/(2a)$

$y = (45 - x)x$

$=> y = -x^2 + 45x + 0$

$=> a = -1$
$=> b = 45$
$=> c = 0$

$x = (-45 +- sqrt(45^2 - 4(-1)(0)))(2(-1))$

$=> x = (-45 +- sqrt(45^2 - 0))/-2$

$=> x = (-45 +- 45)/-2$

$=> x = (-45 + 45)/-2$
$=> x = 0/-2 = 0$

$=> x = (-45 - 45)/-2$
$=> x = -90/-2$
$=> x = 45$

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