Given the equation 2square root of x minus 5 = 2, solve for x and identify if it is an extraneous solution.

Question

x = 6, solution is not extraneous
x = 6, solution is extraneous
x = 11, solution is not extraneous
x = 11, solution is extraneous

Answer 1

To solve the equation \( 2\sqrt{x} - 5 = 2 \), follow these steps:

1. Add \( 5 \) to both sides: \[ 2\sqrt{x} = 7 \]

2. Divide both sides by \( 2 \): \[ \sqrt{x} = \frac{7}{2} \]

3. Square both sides to eliminate the square root: \[ x = \left(\frac{7}{2}\right)^2 = \frac{49}{4} = 12.25 \]

Now, we will check whether this solution is extraneous by substituting \( x = 12.25 \) back into the original equation.

Substituting into \( 2\sqrt{x} - 5 \): \[ 2\sqrt{12.25} - 5 = 2(3.5) - 5 = 7 - 5 = 2 \]

Since both sides of the original equation are equal, we find that the solution is valid and not extraneous.

The final answer is:

• \( x = 12.25 \), solution is not extraneous.

None of the provided options are correct in this case. The solutions given (6 and 11) do not correspond to the solved \( x \) value. If there is a confusion on what options refer to, please clarify or provide the specific problem setup.