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

1. Translate the problem requirements: We have an equation with square roots on both sides: \(\sqrt{3-2\mathrm{x}} = \sqrt{2\mathrm{x}} + 1\). We need to solve for x first, then calculate what \(4\mathrm{x}^2\) equals from the given answer choices.
2. Isolate one square root term: Rearrange the equation to get one square root by itself on one side, making it easier to eliminate the square root by squaring.
3. Square both sides strategically: Square both sides to eliminate one square root, but be prepared to deal with the remaining square root term that emerges.
4. Isolate and eliminate the remaining square root: Get the remaining square root term by itself and square again to completely solve for x.
5. Calculate the target expression: Once we have the value of x, substitute it into \(4\mathrm{x}^2\) to find our answer.

Execution of Strategic Approach

1. Translate the problem requirements

Let's understand what we have and what we need to find. We have an equation with square roots: \(\sqrt{3-2\mathrm{x}} = \sqrt{2\mathrm{x}} + 1\). This means "the square root of (3 minus 2x) equals the square root of 2x plus 1." We need to solve this equation to find the value of x, then calculate what \(4\mathrm{x}^2\) equals.

Before we can work with square roots, we need to consider what values of x are valid. For \(\sqrt{3-2\mathrm{x}}\) to exist, we need \(3-2\mathrm{x} \geq 0\), which means \(\mathrm{x} \leq \frac{3}{2}\). For \(\sqrt{2\mathrm{x}}\) to exist, we need \(2\mathrm{x} \geq 0\), which means \(\mathrm{x} \geq 0\). So our solution must be between 0 and \(\frac{3}{2}\).

Process Skill: TRANSLATE - Converting the square root equation into clear mathematical understanding with domain restrictions

2. Isolate one square root term

Currently we have \(\sqrt{3-2\mathrm{x}} = \sqrt{2\mathrm{x}} + 1\). The left side already has just one square root term by itself, which is perfect for our next step. We want to get one square root isolated so we can eliminate it by squaring.

Our equation is ready: \(\sqrt{3-2\mathrm{x}} = \sqrt{2\mathrm{x}} + 1\)

3. Square both sides strategically

Now we'll square both sides to eliminate the square root on the left. When we square the left side, we get simply \((3-2\mathrm{x})\). When we square the right side, we need to be careful because we have \((\sqrt{2\mathrm{x}} + 1)^2\).

Using the pattern \((\mathrm{a} + \mathrm{b})^2 = \mathrm{a}^2 + 2\mathrm{ab} + \mathrm{b}^2\), where \(\mathrm{a} = \sqrt{2\mathrm{x}}\) and \(\mathrm{b} = 1\):

• \(\mathrm{a}^2 = (\sqrt{2\mathrm{x}})^2 = 2\mathrm{x}\)
• \(2\mathrm{ab} = 2 \times \sqrt{2\mathrm{x}} \times 1 = 2\sqrt{2\mathrm{x}}\)
• \(\mathrm{b}^2 = 1^2 = 1\)

So: \(3 - 2\mathrm{x} = 2\mathrm{x} + 2\sqrt{2\mathrm{x}} + 1\)

Simplifying: \(3 - 2\mathrm{x} = 2\mathrm{x} + 2\sqrt{2\mathrm{x}} + 1\)

4. Isolate and eliminate the remaining square root

Let's rearrange to get the square root term by itself. Starting from: \(3 - 2\mathrm{x} = 2\mathrm{x} + 2\sqrt{2\mathrm{x}} + 1\)

Subtract \(2\mathrm{x}\) and 1 from both sides: \(3 - 2\mathrm{x} - 2\mathrm{x} - 1 = 2\sqrt{2\mathrm{x}}\) \(2 - 4\mathrm{x} = 2\sqrt{2\mathrm{x}}\)

Divide both sides by 2: \(1 - 2\mathrm{x} = \sqrt{2\mathrm{x}}\)

Now square both sides again: \((1 - 2\mathrm{x})^2 = (\sqrt{2\mathrm{x}})^2\)

Expanding the left side: \((1 - 2\mathrm{x})^2 = 1 - 4\mathrm{x} + 4\mathrm{x}^2\) The right side: \((\sqrt{2\mathrm{x}})^2 = 2\mathrm{x}\)

So: \(1 - 4\mathrm{x} + 4\mathrm{x}^2 = 2\mathrm{x}\)

Rearranging: \(4\mathrm{x}^2 - 4\mathrm{x} - 2\mathrm{x} + 1 = 0\) \(4\mathrm{x}^2 - 6\mathrm{x} + 1 = 0\)

Using the quadratic formula: \(\mathrm{x} = \frac{6 \pm \sqrt{36 - 16}}{8} = \frac{6 \pm \sqrt{20}}{8} = \frac{6 \pm 2\sqrt{5}}{8} = \frac{3 \pm \sqrt{5}}{4}\)

We need to check which solution is valid. Since \(1 - 2\mathrm{x} = \sqrt{2\mathrm{x}}\), we need \(1 - 2\mathrm{x} \geq 0\), so \(\mathrm{x} \leq \frac{1}{2}\).

For \(\mathrm{x} = \frac{3 + \sqrt{5}}{4} \approx 1.31\), this is greater than \(\frac{1}{2}\), so it's not valid. For \(\mathrm{x} = \frac{3 - \sqrt{5}}{4} \approx 0.19\), this is less than \(\frac{1}{2}\) and positive, so it's valid.

Therefore: \(\mathrm{x} = \frac{3 - \sqrt{5}}{4}\)

Process Skill: APPLY CONSTRAINTS - Checking which solution satisfies the domain restrictions

5. Calculate the target expression

Now we need to find \(4\mathrm{x}^2\). We have \(\mathrm{x} = \frac{3 - \sqrt{5}}{4}\).

First, let's find \(\mathrm{x}^2\): \(\mathrm{x}^2 = \left(\frac{3 - \sqrt{5}}{4}\right)^2 = \frac{(3 - \sqrt{5})^2}{16} = \frac{9 - 6\sqrt{5} + 5}{16} = \frac{14 - 6\sqrt{5}}{16}\)

Therefore: \(4\mathrm{x}^2 = 4 \times \frac{14 - 6\sqrt{5}}{16} = \frac{14 - 6\sqrt{5}}{4}\)

Let's simplify this differently. From our equation \(4\mathrm{x}^2 - 6\mathrm{x} + 1 = 0\), we can solve for \(4\mathrm{x}^2\): \(4\mathrm{x}^2 = 6\mathrm{x} - 1\)

Since \(\mathrm{x} = \frac{3 - \sqrt{5}}{4}\): \(4\mathrm{x}^2 = 6 \times \frac{3 - \sqrt{5}}{4} - 1 = \frac{6(3 - \sqrt{5})}{4} - 1 = \frac{18 - 6\sqrt{5}}{4} - 1 = \frac{18 - 6\sqrt{5} - 4}{4} = \frac{14 - 6\sqrt{5}}{4}\)

This can be written as: \(4\mathrm{x}^2 = 6\mathrm{x} - 1\)

4. Final Answer

Looking at our answer choices, we found that \(4\mathrm{x}^2 = 6\mathrm{x} - 1\). This matches choice (E) exactly.

Let's verify by checking our original equation with \(\mathrm{x} = \frac{3 - \sqrt{5}}{4}\):

• Left side: \(\sqrt{3 - 2\mathrm{x}} = \sqrt{3 - 2\frac{3-\sqrt{5}}{4}} = \sqrt{\frac{12 - 6 + 2\sqrt{5}}{4}} = \sqrt{\frac{6 + 2\sqrt{5}}{4}}\)
• Right side: \(\sqrt{2\mathrm{x}} + 1 = \sqrt{2\frac{3-\sqrt{5}}{4}} + 1 = \sqrt{\frac{6-2\sqrt{5}}{4}} + 1\)

The algebra confirms our solution is correct.

The answer is (E) \(6\mathrm{x} - 1\).

Common Faltering Points

Errors while devising the approach

1. Ignoring domain restrictions

Students often dive directly into solving the equation without establishing the domain constraints. They miss that for \(\sqrt{3-2\mathrm{x}}\) to exist, we need \(3-2\mathrm{x} \geq 0\) (so \(\mathrm{x} \leq \frac{3}{2}\)), and for \(\sqrt{2\mathrm{x}}\) to exist, we need \(2\mathrm{x} \geq 0\) (so \(\mathrm{x} \geq 0\)). This leads to accepting invalid solutions later.

2. Incorrect squaring strategy

Students may attempt to square both sides immediately without isolating one square root term first. This creates unnecessarily complex expressions and makes the problem much harder to solve.

Errors while executing the approach

1. Expanding (a + b)² incorrectly

When squaring \((\sqrt{2\mathrm{x}} + 1)^2\), students frequently forget the middle term \(2\mathrm{ab}\) and write it as just \((2\mathrm{x}) + 1\) instead of \(2\mathrm{x} + 2\sqrt{2\mathrm{x}} + 1\). This fundamental algebraic error derails the entire solution.

2. Failing to check solution validity

After finding \(\mathrm{x} = \frac{3 \pm \sqrt{5}}{4}\), students often don't verify which solution satisfies the constraint \(1 - 2\mathrm{x} \geq 0\). They may proceed with \(\mathrm{x} = \frac{3 + \sqrt{5}}{4} \approx 1.31\), which violates the requirement that \(\mathrm{x} \leq \frac{1}{2}\).

3. Arithmetic errors during rearrangement

When moving from \(3 - 2\mathrm{x} = 2\mathrm{x} + 2\sqrt{2\mathrm{x}} + 1\) to isolate the square root term, students make sign errors or incorrectly combine like terms, leading to wrong coefficients in the final quadratic equation.

Errors while selecting the answer

1. Providing numerical value instead of expression

Students may calculate the specific numerical value of \(4\mathrm{x}^2 = \frac{14 - 6\sqrt{5}}{4}\) and try to match it to the answer choices, not recognizing that the answer choices are expressions in terms of x and that \(4\mathrm{x}^2 = 6\mathrm{x} - 1\) is the desired form.