How do you solve #10 - x =sqrt(3x + 24)#?

1 Answer
Aug 10, 2015

Answer:

#x = 4#

Explanation:

Right from the start, you know that any solution you find for this equation must satisfy two conditions

  • #3x+24>=0 implies x>=-8#
  • #10-x>=0 implies x <=10#

This means that you must for an #x# that satisfies the overall condition #x in [-8, 10]#.

Since the radical is already isolated on one side of the equation, square both sides to get rid of the square root

#(sqrt(3x + 24))^2 = (10-x)^2#

#3x + 24 = 100 - 20x + x^2#

Rearrange this equation into classic quadratic form

#x^2 -23x +76 = 0#

Use the quadratic formula to find the two solutions to this equation

#x_(1,2) = (-(-23) +- sqrt( (-23)^2 - 4 * 1 * 76))/(2 * 1)#

#x_(1,2) = (23 +- sqrt(225))/2#

#x_(1,2) = (23 +- 15)/2 = {(x_1 = (23 + 15)/2 = 19), (x_2 = (23 - 15)/2 = 4) :}#

Now, since #x=19# #cancel(in) [-8, 10]#, this is not a valid solution solution.

As a result, the only solution to this equation is #x=color(green)(4)#.

Check to see if this is the case

#10 - 4 = sqrt(3 * (4) + 24)#

#6 = sqrt(36)#

#6 = 6# #color(green)(sqrt())#