Question

How do you find the roots, real and imaginary, of `y= (45-x)x- (x-2)(2x-1)` using the quadratic formula?

Answer 1

`x = 25/3+-sqrt(619)/3`

Explanation:

Note that the quadratic equation is given in the form of a function, so the question should really be asking about zeros, not roots.

Proceeding on the assumption that you want to solve for `y=0`...

First we want to get this quadratic into standard form, which we can do by multiplying out and combining terms:

`(45-x)x-(x-2)(2x-1) = (45x-x^2)-(2x^2-5x+2)`

`color(white)((45-x)x-(x-2)(2x-1)) = -3x^2+50x-2`

This is now in the form:

`ax^2+bx+c`

with `a=-3`, `b=50` and `c=-2`

It has zeros given by the quadratic formula:

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

`color(white)(x) = (-(color(blue)(50))+-sqrt((color(blue)(50))^2-4(color(blue)(-3))(color(blue)(-2))))/(2(color(blue)(-3))`

`color(white)(x) = (-50+-sqrt(2500-24))/(-6)`

`color(white)(x) = (50+-sqrt(2476))/6`

`color(white)(x) = (50+-sqrt(2^2 * 619))/6`

`color(white)(x) = (50+-2sqrt(619))/6`

`color(white)(x) = 25/3+-sqrt(619)/3`