Which is the extraneous solution of -x=sq root of 2x+15?
A. -5
B. -3
C. 3
D. 5
I think it is D...?
you are correct
if x = 5
LS = -5
RS = √(10+15) = √25 = 5 ≠ LS
To determine the extraneous solution of the equation -x=sqrt(2x+15), we need to solve the equation and then check if any solutions are extraneous.
Step 1: Start with the equation -x=sqrt(2x+15).
Step 2: Square both sides of the equation to eliminate the square root: (-x)^2 = (sqrt(2x+15))^2.
This simplifies to x^2 = 2x + 15.
Step 3: Rearrange the equation to bring all terms to one side: x^2 - 2x - 15 = 0.
Step 4: Factorize the equation (if possible) or use the quadratic formula to solve for x.
Factoring the equation: (x - 5)(x + 3) = 0.
Setting each factor equal to zero, we have x - 5 = 0, and x + 3 = 0.
Solving these equations, we get x = 5 and x = -3.
Step 5: Check each solution in the original equation.
For x = 5: -5 = sqrt(2(5) + 15) which simplifies to -5 = sqrt(25), which is true.
For x = -3: -(-3) = sqrt(2(-3) + 15) which simplifies to 3 = sqrt(9), which is true.
Step 6: Since both solutions fit the original equation, there are no extraneous solutions. Therefore, the correct answer is D) 5, as it is not extraneous.
To find the extraneous solution of the equation -x = √(2x + 15), we need to solve the equation and then check if any of the solutions result in an invalid answer.
First, let's solve the equation:
-x = √(2x + 15)
To eliminate the square root, we can square both sides of the equation:
(-x)^2 = (√(2x + 15))^2
x^2 = 2x + 15
Moving all the terms to one side, we get:
x^2 - 2x - 15 = 0
Now, we can solve this quadratic equation by factoring or using the quadratic formula:
(x - 5)(x + 3) = 0
Setting each factor equal to zero:
x - 5 = 0 or x + 3 = 0
Solving for x, we find two potential solutions:
x = 5 or x = -3
Now, let's check each of these solutions to see if they satisfy the original equation -x = √(2x + 15):
For x = 5:
-x = -5
√(2(5) + 15) = √(25) = 5, which is true.
For x = -3:
-x = -(-3) = 3
√(2(-3) + 15) = √(9) = 3, which is also true.
Both x = 5 and x = -3 satisfy the original equation, so neither of them is an extraneous solution.
Therefore, the correct answer is: None of the provided options (A. -5, B. -3, C. 3, D. 5) is an extraneous solution.