How do you solve #sqrt(4x+1) - sqrt(x-2)=3#?

1 Answer
Aug 11, 2015

Answer:

#{(x = 6), (x=2) :}#

Explanation:

Start by writing the conditions that a value of #x# must satisfy in order to be a valid solution to this equation.

You have to do that because you're dealing with radical terms, which means that the expressions under the radical signs must be positive for real numbers.

  • #4x+1>=0 implies x >=-1/4#
  • #x-2>=0 implies x >= 2#

Combine these two conditions to get #x>=2#.

Now go on to solve the equation. Square both sides to get

#(sqrt(4x+1) - sqrt(x-2))^2 = 3^2#

#(sqrt(4x+1))^2 - 2sqrt((4x+1)(x-2)) + (sqrt(x-2))^2 = 9#

#4x + 1 - 2sqrt((4x+1)(x-2)) + x - 2 = 9#

This is equivalent to

#2sqrt((4x+1)(x-2)) = 5x - 10#

Square both sides of the equation again to get rid of the last radical term

#(2sqrt((4x+1)(x-2)))^2 = (5x-10)^2#

#4(4x+1)(x-2) = 25x^2 - 100x + 100#

#16x^2 - 28x -8 = 25x^2 - 100x + 100#

#9x^2 = 72x + 108 = 0#

Simplify this quadratic equation by dividing everything by #9#

#(color(red)(cancel(color(black)(9)))x^2)/color(red)(cancel(color(black)(9))) - 8x + 12 = 0#

You can find the two roots to this quadratic by using the quadratic formula

#x_(1,2) = (-(8) +- sqrt( (-8)^2 - 4 * 1 * 12))/(2 * 1)#

#x_(1,2) = (8 +- sqrt(16))/2 = (8 +- 4)/2 = {(x_1 = (8 + 4)/2 = 6), (x_2 = (8-4)/2 = 2) :}#

Since both #x_1# and #x_2# satisfy your condition #x>=2#, they will both be valid solutions to the original equation.

You can do a quick check to make sure that you got the calculations right

#sqrt(4 * 6 + 1) - sqrt(6-2) = 3#

#sqrt(25) - sqrt(4) = 3#

#5 - 2 = 3 color(green)(sqrt())#

and

#sqrt(4 * 2 + 1) - sqrt(2-2) = 3#

#sqrt(9) - sqrt(0) = 3#

#3 - 0 = 3 color(green)(sqrt())#