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

Question

Julia Adams · Accepted Answer

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

#=(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#.

Savannah Anderson · Answer

undefined To find the roots of the quadratic equation $y = \left(\frac{x}{5} - 5\right)\left(-\frac{x}{2} - 2\right)$ using the quadratic formula, follow these steps:

1. Expand the expression to get a quadratic equation in standard form.
2. Identify the coefficients $a$, $b$, and $c$.
3. Substitute the values of $a$, $b$, and $c$ into the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
4. Calculate the discriminant $b^2 - 4ac$.
5. If the discriminant is positive, the roots are real and distinct. If it's zero, the roots are real and equal. If it's negative, the roots are imaginary.
6. Use the quadratic formula to find the roots, substituting the values of $a$, $b$, and $c$, and then solve for $x$.