what is the solution of sqrt(2x+4)-sqrt(x)=2 A. x = 0 B. x = 0 and x = 16 C. x = 0 and x = –16 D. x = 16 and x = –16
2 answers:
We want to solve
√(2x+4) - √(x) = 2
Write equation as
√(2x+4) = √x + 2
Square each side.
2x + 4 = x + 4√x + 4
x = 4√x
x - 4√x = 0
√x (√x - 4) = 0
Either
√x = 0 => x = 0
or
√x = 4 => x = 16
Test for extraneous solutions.
When x = 0:
√(2x+4) - √x = 2 (Correct)
When x = 16:
√(2x+4) - √x = √(36) - √(16) = 6 - 4 = 2 (Correct)
A plot of f(x) = √(2x+4) - √x - 2 = 0 confirms hat the solutions are correct.
Answer: B. x = 0 and x = 16.
√(2x+4) - √(x) = 2
Write equation as
√(2x+4) = √x + 2
Square each side.
2x + 4 = x + 4√x + 4
x = 4√x
x - 4√x = 0
√x (√x - 4) = 0
Either
√x = 0 => x = 0
or
√x = 4 => x = 16
Test for extraneous solutions.
When x = 0:
√(2x+4) - √x = 2 (Correct)
When x = 16:
√(2x+4) - √x = √(36) - √(16) = 6 - 4 = 2 (Correct)
A plot of f(x) = √(2x+4) - √x - 2 = 0 confirms hat the solutions are correct.
Answer: B. x = 0 and x = 16.
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Answer:
Tristan is correct, because he needs to find the possible dimensions for rectangle WXYZ and then divide each dimension by 3.
Step-by-step explanation:
The rectangle ABCD is dilated by a factor of 3 to get the rectangle WXYZ whose area is found to be 72 cm².
Let the dimensions of the dilated rectangle WXYZ are a cm by b cm.
So, ab = 72 ........... (1)
Now, the dimensions of the original rectangle are 3 times lesser than the dimensions of rectangle WXYZ.
So, the area of rectangle ABCD will be cm² {from equation (1)}
Therefore, Tristan is correct, because he needs to find the possible dimensions for rectangle WXYZ and then divide each dimension by 3. (Answer)