How do you solve #4=x+2sqrt(x-7)# and identify any restrictions?

1 Answer
Mar 24, 2017

Answer:

#x = 6 +2i sqrt2, 6 -2i sqrt2#
There is no real root

Explanation:

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

square both sides
#(4 -x)^2 = (2 sqrt(x-7))^2#

#16 - 8 x +x^2 = 4(x-7)#
#16 - 8 x +x^2 = 4x- 28#

rearrange the equation,
#x^2 - 8x - 4x + 16 + 28 = 0#
#x^2 - 12x + 44 = 0#

#a = 1, b = -12, c =44#
since #b^2 - 4ac < 0#, so there is no real root for this equation.

We use completing a square to solve it.
#x^2 - 12x + 44 = 0#
#(x -6)^2 - (-6)^2 + 44 = 0#
#(x -6)^2 - 36 + 44 = 0#
#(x -6)^2 + 8 = 0#
#(x -6)^2 = -8#
#(x -6) = +-sqrt (-8) = +-sqrt (-1 * 8) = +- i sqrt 8 = +-2i sqrt2#
#x = 6 +-2i sqrt2#