How do you find the roots, real and imaginary, of #y=(x/5-5)(-3x-2)# using the quadratic formula?

1 Answer
Jul 24, 2017

Answer:

See a solution process below:

Explanation:

First, we need to write this equation in standard form:

#y = (color(red)(x/5) - color(red)(5))(color(blue)(-3x) - color(blue)(2))# becomes:

#y = -(color(red)(x/5) xx color(blue)(3x)) - (color(red)(x/5) xx color(blue)(2)) + (color(red)(5) xx color(blue)(3x)) + (color(red)(5) xx color(blue)(2))#

#y = -3/5x^2 - 2/5x + 15x + 10#

#y = -3/5x^2 - 2/5x + (5/5 * 15)x + 10#

#y = -3/5x^2 - 2/5x + 75/5x + 10#

#y = -3/5x^2 + 73/5x + 10#

We can now use the quadratic equation to solve the equation. The quadratic formula states:

For #color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0#, the values of #x# which are the solutions to the equation are given by:

#x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - 4color(red)(a)color(green)(c)))/(2color(red)(a))#

Substituting:

#color(red)(-3/5)# for #color(red)(a)#

#color(blue)(73/5)# for #color(blue)(b)#

#color(green)(10)# for #color(green)(c)# gives:

#x = (-color(blue)(73/5) +- sqrt(color(blue)(73/5)^2 - (4 * color(red)(-3/5) * color(green)(10))))/(2 * color(red)(-3/5))#

#x = (-color(blue)(73/5) +- sqrt(5329/25 - (-120/5)))/(-6/5)#

#x = (-color(blue)(73/5) +- sqrt(5329/25 - (5/5 * -120/5)))/(-6/5)#

#x = (-color(blue)(73/5) +- sqrt(5329/25 - (-600/25)))/(-6/5)#

#x = (-color(blue)(73/5) +- sqrt(5329/25 + 600/25))/(-6/5)#

#x = (-color(blue)(73/5) +- sqrt(5929/25))/(-6/5)#

#x = (-color(blue)(73/5) + 77/5)/(-6/5)# and #x = (-color(blue)(73/5) - 77/5)/(-6/5)#

#x = (4/5)/(-6/5)# and #x = (-150/5)/(-6/5)#

#x = -(4 * 5)/(5 * 6)# and #x = (150 * 5)/(5 * 6)#

#x = -4/6# and #x = (150)/(6)#

#x = -2/3# and #x = 25#