Answer: x = 8
x = 1 is an extraneous solution
Explanation:
- x + 3 = 0
RESTRICTIONS: 3x + 1 ≥ 0 → x ≥
(because square roots cannot be less than zero/negative)
- x + 3 = 0
+x -3 +x -3
= x - 3
= (x - 3)²
3x + 1 = x² - 6x + 9
-3x -1 -3x -1
0 = x² - 9x + 8
0 = (x - 1)(x - 8)
0 = x - 1 0 = x - 8
+1 +1 +8 +8
1 = x 8 = x both solutions are ≥
Check:
x = 1:
- (1) + 3 = 0
2 - 1 + 3 = 0
4 = 0
FALSE so x = 1 is NOT a valid solution
x = 8:
- (8) + 3 = 0
5 - 8 + 3 = 0
0 = 0
TRUE so x = 8 is a valid solution