Question

solve the equation.
x - \sqrt{3 - 2x} = 0

select the correct choice below and, if necessary, fill in the answer box to complete your choice.
a. the solution set is {}.
(simplify your answer. use a comma to separate answers as needed.)
b. there is no solution.

Explanation:

Step1: Isolate the square root

We start with the equation \( x - \sqrt{3 - 2x} = 0 \). To isolate the square root, we add \( \sqrt{3 - 2x} \) to both sides and subtract \( 0 \) (which doesn't change the equation), getting \( x = \sqrt{3 - 2x} \).

Step2: Square both sides

To eliminate the square root, we square both sides of the equation. Squaring the left side \( x \) gives \( x^2 \), and squaring the right side \( \sqrt{3 - 2x} \) gives \( 3 - 2x \). So we have the quadratic equation \( x^2 = 3 - 2x \).

Step3: Rearrange into standard quadratic form

We rearrange the equation \( x^2 = 3 - 2x \) to get it in standard quadratic form \( ax^2 + bx + c = 0 \). Adding \( 2x \) and subtracting \( 3 \) from both sides, we obtain \( x^2 + 2x - 3 = 0 \).

Step4: Factor the quadratic equation

We factor the quadratic \( x^2 + 2x - 3 \). We need two numbers that multiply to \( -3 \) and add to \( 2 \). The numbers \( 3 \) and \( -1 \) work, so we can factor it as \( (x + 3)(x - 1) = 0 \).

Step5: Solve for x

Setting each factor equal to zero gives us two solutions:

• For \( x + 3 = 0 \), we subtract \( 3 \) from both sides to get \( x = -3 \).
• For \( x - 1 = 0 \), we add \( 1 \) to both sides to get \( x = 1 \).

Step6: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.

• For \( x = -3 \): Substitute into the original equation \( x - \sqrt{3 - 2x} \). First, calculate the square root part: \( 3 - 2(-3) = 3 + 6 = 9 \), so \( \sqrt{9} = 3 \). Then, \( -3 - 3 = -6

eq 0 \), so \( x = -3 \) is not a solution.

• For \( x = 1 \): Substitute into the original equation. Calculate the square root part: \( 3 - 2(1) = 3 - 2 = 1 \), so \( \sqrt{1} = 1 \). Then, \( 1 - 1 = 0 \), which satisfies the original equation. So \( x = 1 \) is a valid solution.

Answer:

The solution set is \(\{1\}\)