How do you solve #sqrt(4x+17) +sqrt(x +7) =sqrt(x + 2)#?
1 Answer
Explanation:
Start by writing down the conditions that
#4x + 17>=0 implies x>= -17/4# #x+7 >=0 implies x>= -7# #x+2 >=0 implies x>= -2#
These conditions are determined by the fact that the expressions under the radicals must be positive if you're dealing with real numbers.
Combine all three conditions to get
Another important thing to notice here is that for any
#4x+17 >x+7#
Moreover,
#x+7 > x+2#
This actually tells you that the equation has no valid solutions among real numbers, since two positive numbers cannot be added to give a smaller positive number.
In other simply, any solution that you will come about by solving this equation will be extraneous.
Now, square both sides of the equation to reduce the number of radical terms from three to one.
#(sqrt(4x+17) + sqrt(x+7))^2 = (sqrt(x+2))^2#
#(sqrt(4x+17))^2 + 2sqrt((4x+17)(x+7)) + (sqrt(x+7))^2 = x+2#
#4x+17 + 2sqrt((4x+17)(x+7)) + color(red)(cancel(color(black)(x))) + 7= color(red)(cancel(color(black)(x))) + 2#
This is equivalent to
#2sqrt((4x+17)(x+7)) = -4x - 22#
Square both sides of the equation again to get rid of the last radical term
#(2sqrt((4x+17)(x+7)))^2 = (-4x - 22)""^2#
#4(4x+17)(x+7) = 16x^2 + 176x + 484#
#color(red)(cancel(color(black)(16x^2))) + 180x + 476 = color(red)(cancel(color(black)(16x^2))) + 176x + 484#
This is equivalent to
#4x = 8 implies x = 8/4 = color(green)(2)#
Notice that
#sqrt(4 * (2) + 17) + sqrt(2 + 7) = sqrt(2 + 2)#
#sqrt(25) + sqrt(9) = sqrt(4)#
#5 + 3 color(Red)(!=) 2 -> x = 2# is an extraneous solution.