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

Question

Eli Assouline · Accepted Answer

See a solution process below:

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#

Nathan Axel · Answer

undefined To find the roots, real and imaginary, of $ y = \left(\frac{x}{5} - 5\right)(-3x - 2) $ using the quadratic formula, we first expand and simplify the equation to get it in standard quadratic form, $ ax^2 + bx + c = 0 $.

Expanding and simplifying the given equation:

$$ y = \left(\frac{x}{5} - 5\right)(-3x - 2) $$
$$ y = -\frac{3x^2}{5} + 15x + 10 $$

Now, we have $ a = -\frac{3}{5} $, $ b = 15 $, and $ c = 10 $.

Using the quadratic formula:

$$ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$

Substitute the values of $ a $, $ b $, and $ c $ into the formula:

$$ x = \frac{-15 \pm \sqrt{15^2 - 4(-\frac{3}{5})(10)}}{2(-\frac{3}{5})} $$

Simplify and solve:

$$ x = \frac{-15 \pm \sqrt{225 + 24}}{-\frac{6}{5}} $$
$$ x = \frac{-15 \pm \sqrt{249}}{-\frac{6}{5}} $$

Therefore, the roots of the equation are:

$$ x = \frac{15 \pm \sqrt{249}}{\frac{6}{5}} $$
$$ x = \frac{25}{3} \pm \frac{5\sqrt{249}}{3} $$

These are the real and imaginary roots of the given quadratic equation.