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

##### 1 Answer

#### Answer:

#### Explanation:

Start by writing the conditions that a value of

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

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

#(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

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())#