Answer:
x = 10
Explanation:
I assume you mean
sqrt(x + 6) = x - 6
let's square the whole equation :
x + 6 = (x - 6)² = x² - 12x + 36
x² - 13x + 30 = 0
the general solution to a quadratic equation is
x = (-b ±sqrt(b² - 4ac))/(2a)
in our case
a = 1
b = -13
c = 30
x = (13 ±sqrt(13² - 4×1×30))/(2×1) =
= (13 ±sqrt(169 - 120))/2 = (13 ±sqrt(49))/2 =
= (13 ± 7)/2
x1 = (13 + 7)/2 = 20/2 = 10
x2 = (13 - 7)/2 = 6/2 = 3
for x = 10 we get for the original equation
sqrt(10+6) = 10-6 = 4
sqrt(16) = 4
4 = 4
so, x = 10 is a valid solution.
for x = 3 we get
sqrt(3 + 6) = 3 - 6 = -3
sqrt(9) = -3
3 = -3
so, this (x = 3) counts as an invalid solution, because most teachers nowadays expect the square root symbol operation to deliver only the positive solution.
although a square root is actually representing both : the positive AND the negative solution, because the negative number multiplied by itself is the same result as the positive number multiplied by itself.
so, actually
sqrt(9) = -3
± 3 = -3
is still correct and valid. but not what your teacher expects most likely.
in my younger days only the square root of a negative number counted as invalid operation (in the space of Real numbers, of course), and therefore only a solution leading to a square root of a negative number was an invalid solution.