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

1 Answer
May 1, 2018

Answer:

#x_1 = 25 and x_2 = -4# are the two real roots of the given function.

Explanation:

#=(x/5)(-x/2 -2) - 5(-x/2 -2)#
#=-x^2/10 -(2x)/5 + (5x)/2 + 10#
#= (-x^2 - 4x + 25x + 100)/10#

Note that by definition, the roots equal 0. So we can set #y = 0#
#implies 0 = (-x^2 - 4x + 25x + 100)/10#
#implies (-x^2 - 4x + 25x + 100)= 0#
Let's multiply this by -1 to simply things.
#(x^2 + 4x - 25x - 100) = 0#
Funnily, you can quite easily factorize this polynomial: by factoring x and -25. However, you need to use the formula, which complicates things.
#(x^2- 21x - 100) = 0# The formula is #x_1 = [-b + sqrt(b^2 - 4ac)]/(2a)#
#x_2 = [-b - sqrt(b^2 - 4ac)]/(2a)#

Substitute #a = 1, b = -21 and c= -100# into both of these to get the two roots. I will not solve it here as it would clutter the answer, however, the end result would be #x_1 = 25 and x_2 = -4#.