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

1 Answer
George C.
Feb 24, 2018

#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#

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