Question

solve the radical equation.

\\\sqrt{11-x} - x = 1\\

select the correct choice below and, if necessary, fill in the answer box to complete your choice.

a. the solution set is \\{\quad\\}. (use a comma to separate answers as needed.)
b. the solution set is the empty set.

Explanation:

🆕 New Concept Discovered: Solving Radical Equations
Isolate the radical, square both sides, and check for extraneous solutions.

Step 1: Isolate the radical term

To solve the equation, we first want to get the square root term by itself on one side. We do this by adding \( x \) to both sides of the equation:

$$ \sqrt{11 - x} = x + 1 $$

Step 2: Square both sides

To eliminate the radical, we square both sides of the equation. Remember that squaring a binomial like \( (x + 1) \) requires expanding it:

$$ (\sqrt{11 - x})^2 = (x + 1)^2 $$

$$ 11 - x = x^2 + 2x + 1 $$

Step 3: Set the quadratic equation to zero

Move all terms to one side to set the quadratic equation to zero:

$$ 0 = x^2 + 2x + x + 1 - 11 $$

$$ x^2 + 3x - 10 = 0 $$

Step 4: Factor and solve the quadratic equation

We look for two numbers that multiply to \( -10 \) and add to \( 3 \). These numbers are \( 5 \) and \( -2 \):

$$ (x + 5)(x - 2) = 0 $$

This gives two potential solutions:

$$ x = -5 \quad \text{or} \quad x = 2 $$

Step 5: Check for extraneous solutions

Because we squared both sides of the equation, we must plug our potential solutions back into the original equation, \( \sqrt{11 - x} - x = 1 \), to see if they are valid.

• Test \( x = 2 \):

$$ \sqrt{11 - 2} - 2 = \sqrt{9} - 2 = 3 - 2 = 1 $$

Since \( 1 = 1 \), \( x = 2 \) is a valid solution.

• Test \( x = -5 \):

$$ \sqrt{11 - (-5)} - (-5) = \sqrt{16} + 5 = 4 + 5 = 9 $$

Since \( 9
eq 1 \), \( x = -5 \) is an extraneous solution and must be discarded.

Answer:

A. The solution set is \( \{2\} \).