How do you solve #[(x^2+5x) / (x +2)] + 10x = -24#?

1 Answer
Nov 5, 2017

See a solution process below:

Explanation:

First, subtract #color(red)(10x)# from each side of the equation to isolate the fraction while keeping the equation balanced:

#[(x^2 + 5x)/(x + 2)] + 10x - color(red)(10x) = -24 - color(red)(10x)#

#[(x^2 + 5x)/(x + 2)] + 0 = -24 - 10x#

#[(x^2 + 5x)/(x + 2)] = -10x - 24#

Next, multiply each side of the equation by #color(red)((x + 2))# to eliminate the fraction while keeping the equation balanced:

#color(red)((x + 2))[(x^2 + 5x)/(x + 2)] = color(red)((x + 2))(-10x - 24)#

#cancel(color(red)((x + 2)))[(x^2 + 5x)/color(red)(cancel(color(black)(x + 2)))] = (color(red)(x) * -10x) + (color(red)(x) * -24) + (color(red)(2) * -10x) + (color(red)(2) * -24)#

#x^2 + 5x = -10x^2 - 24x - 20x - 48#

#x^2 + 5x = -10x^2 - 44x - 48#

Then, add #color(red)(10x^2)# and #color(blue)(44x)# and #color(orange)(48)# to each side of the equation to put the equation in standard form:

#x^2 + color(red)(10x^2) + 5x + color(blue)(44x) + color(orange)(48) = -10x^2 + color(red)(10x^2) - 44x + color(blue)(44x) - 48 + color(orange)(48)#

#1x^2 + color(red)(10x^2) + 5x + color(blue)(44x) + color(orange)(48) = 0 - 0 - 0#

#(1 + color(red)(10))x^2 + (5 + color(blue)(44))x + 48 = 0#

#11x^2 + 49x + 48 = 0#

We can now use the quadratic equation to solve this problem:

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))))/(2 * color(red)(a))#

Substituting:

#color(red)(11)# for #color(red)(a)#

#color(blue)(49)# for #color(blue)(b)#

#color(green)(48)# for #color(green)(c)# gives:

#x = (-color(blue)(49) +- sqrt(color(blue)(49)^2 - (4 * color(red)(11) * color(green)(48))))/(2 * color(red)(11)#

#x = (-color(blue)(49) +- sqrt(2401 - 2112))/22#

#x = (-color(blue)(49) +- sqrt(289))/22#

#x = (-color(blue)(49) - 17)/22# and #x = (-color(blue)(49) + 17)/22#

#x = -66/22# and #x = -32/22#

#x = -3# and #x = -16/11#