Question

If \(\sqrt{3-2x} = \sqrt{2x + 1}\), then \(4x^2 =\)

• A. 1
• B. 4
• C. 2 -- 2x
• D. 4x -- 2
• E. 6x -- 1

Hint:

When solving radical equations, always remember to check your solutions by plugging them back into the original equation. Squaring both sides can sometimes introduce extraneous solutions that are not valid. Also, be sure to answer what the question is asking for (e.g., \(4x^2\), not just \(x\)).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We are given an equation involving square roots and need to solve for an expression involving the variable x. The key first step is to eliminate the square roots by squaring both sides of the equation.
Step 2: Key Formula or Approach:
Assuming the equation is \(\sqrt{3-2x} = \sqrt{2x+1}\).
To solve, we will square both sides: \[ (\sqrt{3-2x})^2 = (\sqrt{2x+1})^2 \] This simplifies the equation to a linear one, which we can then solve for x.
Step 3: Detailed Explanation:
Squaring both sides of the equation gives: \[ 3 - 2x = 2x + 1 \] Now, solve for x. Add 2x to both sides: \[ 3 = 4x + 1 \] Subtract 1 from both sides: \[ 2 = 4x \] Divide by 4: \[ x = \frac{2}{4} = \frac{1}{2} \] The problem asks for the value of \(4x^2\), not x. Substitute the value of x we found: \[ 4x^2 = 4\left(\frac{1}{2}\right)^2 = 4\left(\frac{1}{4}\right) = 1 \] We should also check if the solution \(x = 1/2\) is valid by plugging it back into the original equation's radicals. For \(\sqrt{3-2x}\): \(\sqrt{3 - 2(1/2)} = \sqrt{3-1} = \sqrt{2}\). (Valid)
For \(\sqrt{2x+1}\): \(\sqrt{2(1/2)+1} = \sqrt{1+1} = \sqrt{2}\). (Valid)
Since \(\sqrt{2} = \sqrt{2}\), the solution is correct.
Step 4: Final Answer:
The value of \(4x^2\) is 1.