How do you solve # (sqrt(x + 7)) - 2(sqrt(x)) = -2# and find any extraneous solutions?

1 Answer
Aug 4, 2016

Answer:

#x=9#

Explanation:

The first thing to do here is write out the conditions that a value of #x# must satisfy in order to be a valid solution to the equation

#sqrt(x+7) - 2sqrt(x) = -2#

You're working with real numbers, so right from the start you know that

  • you can only take the square root of a positive number
  • the square root of a positive number will always return a positive number

In your case, this means that you must have

#x + 7 >= 0 implies x >= -7#

#x >= 0#

Combine these two conditions to get

#x in [-7, + oo) nn [0, + oo) implies color(purple)(|bar(ul(color(white)(a/a)color(black)(x in [0, + oo))color(white)(a/a)|)))#

Now, your goal here is to get rid of the radical terms. To do that, square both sides of the equation

#(sqrt(x+7) - 2sqrt(x))^2 = (-2)^2#

This will get you

#x + 7 - 4 * sqrt((x+7) * x) + 4x = 4#

Isolate the remaining radical term on one side of the equation

#-4 * sqrt( (x+7) * x) = -5x - 3#

#4 * sqrt((x+7) + x) = 5x + 3#

Square both sides of the equation again

#(4 * sqrt((x+7) * x))^2 = (5x + 3)^2#

#16 * (x^2 + 7x) = 25x^2 + 30x + 9#

Rearrange to quadratic form

#9x^2 - 82x + 9 = 0#

Use the quadratic formula to find the two roots

#x_(1,2) = (-(82) +- sqrt( (-82)^2 - 4 * 9 * 9))/(2 * 9)#

#x_(1,2) = (82 +- sqrt(6400))/18#

#x_(1,2) = (82 +- 80)/18 implies {(x_1 = (82 + 80)/18 = 9), (x_2 = (82-80)/18 = 1/9) :}#

Since you have

#x_1, x_2 in [0, +oo)#

both roots can be valid solutions to the original equation. Do a quick calculation to see if both roots are indeed valid solutions

#sqrt(9 + 7) - 2 * sqrt(9) = -2#

#4 - 2 * 3 = -2 " "color(green)(sqrt())#

#sqrt(1/9 + 7) - 2 * sqrt(1/9) = -2#

#sqrt(64/9) - 2 * 1/3 = -2#

#8/3 - 2 * 1/3 = -2 " "color(red)(xx)#

As you can see, #x = 1/9# is not a solution to the original equation, i.e. it's an extraneous solution. Therefore, you can say that the original equation has one valid solution, #x=9#.