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

1 Answer
Karthik N
May 1, 2018

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.

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