How do you solve # (sqrt(x + 7)) - 2(sqrt(x)) = -2# and find any extraneous solutions?
1 Answer
Explanation:
The first thing to do here is write out the conditions that a value of
#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,