Answer:
3. x = 4, NOT extraneous
4. x=12, extraneous
Step-by-step explanation:
3. Solve for x, √(2x +1) = 3 and identify if it's an extraneous solution or not.
Since the square root is already isolated on its side, the first step is to put both sides to their square value to get rid of the square root. We then get:
2x + 1 = 9 => 2x = 8 => x = 4
To see if x = 4 is an extraneous solution, we enter it in the original equation and see if it works:
√(2x +1) = √(2 * 4 +1) = √9 = 3
3 equals 3... so it's NOT an extraneous solution
4. Solve for x, -4√(x-3) = 12 and identify if it's an extraneous solution or not.
This time the square root isn't alone on its side yet... so we have to isolate it, by dividing each side by -4:
-4√(x-3) = 12 becomes √(x-3) = -3
Then we raise each side to their square value to get rid of the square root:
x - 3 = 9 which becomes x = 12
To see if x = 12 is an extraneous solution, we enter it in the original equation and see if it works:
-4√(x-3) = -4√(12-3) = -4√(9) = -4 (3) = -12
-12 does not equal 12... so it's an extraneous solution
