If x is a real number and x+sqrt(x)=1, which of the following is the value of sqrt(x)?

1. Translate the problem requirements: We have an equation \(\mathrm{x} + \sqrt{\mathrm{x}} = 1\) where x is a real number, and we need to find the value of \(\sqrt{\mathrm{x}}\) from the given answer choices.
2. Set up substitution to simplify: Let \(\mathrm{y} = \sqrt{\mathrm{x}}\) to convert the equation into a more manageable quadratic form, since this eliminates the square root complexity.
3. Solve the resulting quadratic equation: Use the quadratic formula to find the possible values of y, keeping in mind that \(\mathrm{y} = \sqrt{\mathrm{x}}\) must be non-negative.
4. Verify and select the valid solution: Check which solution satisfies the constraint that \(\mathrm{y} \geq 0\) and matches one of the given answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

We have an equation where some number x plus the square root of that same number equals 1. In mathematical terms, this is \(\mathrm{x} + \sqrt{\mathrm{x}} = 1\). We need to find what \(\sqrt{\mathrm{x}}\) equals from the given answer choices.

The key insight here is that we're not looking for x itself, but rather for \(\sqrt{\mathrm{x}}\). This suggests we should think about \(\sqrt{\mathrm{x}}\) as our main unknown rather than x.

Process Skill: TRANSLATE - Converting the problem statement into a clear mathematical understanding of what we need to find

2. Set up substitution to simplify

Instead of working directly with the square root, let's make our life easier by giving \(\sqrt{\mathrm{x}}\) a simple name. Let's call \(\sqrt{\mathrm{x}} = \mathrm{y}\).

If \(\sqrt{\mathrm{x}} = \mathrm{y}\), then what is x? Well, if we square both sides of \(\sqrt{\mathrm{x}} = \mathrm{y}\), we get \(\mathrm{x} = \mathrm{y}^2\).

Now our original equation \(\mathrm{x} + \sqrt{\mathrm{x}} = 1\) becomes much cleaner:
\(\mathrm{y}^2 + \mathrm{y} = 1\)

This is now a simple quadratic equation! We've transformed a problem with square roots into a straightforward algebra problem.

Process Skill: SIMPLIFY - Using substitution to eliminate the complexity of square roots

3. Solve the resulting quadratic equation

We have \(\mathrm{y}^2 + \mathrm{y} = 1\). Let's rearrange this into standard form:
\(\mathrm{y}^2 + \mathrm{y} - 1 = 0\)

Using the quadratic formula where a = 1, b = 1, and c = -1:
\(\mathrm{y} = \frac{-\mathrm{b} \pm \sqrt{\mathrm{b}^2 - 4\mathrm{ac}}}{2\mathrm{a}}\)
\(\mathrm{y} = \frac{-1 \pm \sqrt{1^2 - 4(1)(-1)}}{2(1)}\)
\(\mathrm{y} = \frac{-1 \pm \sqrt{1 + 4}}{2}\)
\(\mathrm{y} = \frac{-1 \pm \sqrt{5}}{2}\)

This gives us two solutions:
\(\mathrm{y} = \frac{-1 + \sqrt{5}}{2}\) or \(\mathrm{y} = \frac{-1 - \sqrt{5}}{2}\)

We can rewrite these as:
\(\mathrm{y} = \frac{\sqrt{5} - 1}{2}\) or \(\mathrm{y} = -\frac{\sqrt{5} + 1}{2}\)

4. Verify and select the valid solution

Remember that \(\mathrm{y} = \sqrt{\mathrm{x}}\), and square roots of real numbers must be non-negative (zero or positive).

Let's check our two solutions:

• \(\mathrm{y} = \frac{\sqrt{5} - 1}{2}\): Since \(\sqrt{5} \approx 2.236\), this gives us \(\frac{2.236 - 1}{2} = \frac{1.236}{2} \approx 0.618\), which is positive ✓
• \(\mathrm{y} = -\frac{\sqrt{5} + 1}{2}\): This is clearly negative since \(\sqrt{5} + 1 > 0\), so the negative sign makes the whole expression negative ✗

Therefore, \(\sqrt{\mathrm{x}} = \frac{\sqrt{5} - 1}{2}\)

Looking at our answer choices, this matches choice A: \(\frac{1}{2}(\sqrt{5} - 1)\)

Process Skill: APPLY CONSTRAINTS - Recognizing that \(\sqrt{\mathrm{x}}\) must be non-negative to eliminate invalid solutions

Final Answer

The value of \(\sqrt{\mathrm{x}}\) is \(\frac{1}{2}(\sqrt{5} - 1)\), which corresponds to answer choice A.

We can verify: If \(\sqrt{\mathrm{x}} = \frac{\sqrt{5} - 1}{2}\), then \(\mathrm{x} = \left(\frac{\sqrt{5} - 1}{2}\right)^2 = \frac{5 - 2\sqrt{5} + 1}{4} = \frac{6 - 2\sqrt{5}}{4} = \frac{3 - \sqrt{5}}{2}\)
Checking: \(\mathrm{x} + \sqrt{\mathrm{x}} = \frac{3 - \sqrt{5}}{2} + \frac{\sqrt{5} - 1}{2} = \frac{3 - \sqrt{5} + \sqrt{5} - 1}{2} = \frac{2}{2} = 1\) ✓

Common Faltering Points

Errors while devising the approach

1. Attempting to solve for x directly instead of √x
Many students see the equation \(\mathrm{x} + \sqrt{\mathrm{x}} = 1\) and immediately try to isolate x by moving \(\sqrt{\mathrm{x}}\) to the other side, getting \(\mathrm{x} = 1 - \sqrt{\mathrm{x}}\). They then get stuck because they still have both x and \(\sqrt{\mathrm{x}}\) in their equation. The key insight is recognizing that we need \(\sqrt{\mathrm{x}}\) as our answer, so we should make \(\sqrt{\mathrm{x}}\) our primary variable through substitution.

2. Missing the substitution strategy
Students often try to work directly with the square root expression \(\sqrt{\mathrm{x}}\) throughout their solution, making the algebra much more complex than necessary. The strategic move is to substitute \(\mathrm{y} = \sqrt{\mathrm{x}}\), which transforms the equation into the much simpler \(\mathrm{y}^2 + \mathrm{y} = 1\). Without this substitution, students get bogged down in complicated square root manipulations.

Errors while executing the approach

1. Quadratic formula calculation errors
When applying the quadratic formula to \(\mathrm{y}^2 + \mathrm{y} - 1 = 0\), students commonly make arithmetic mistakes. The most frequent error is incorrectly calculating the discriminant: they might compute \(\mathrm{b}^2 - 4\mathrm{ac}\) as \(1 - 4 = -3\) instead of the correct \(1 + 4 = 5\). This leads to an imaginary result instead of the correct \(\pm\sqrt{5}\).

2. Sign errors when rearranging solutions
After getting \(\mathrm{y} = \frac{-1 \pm \sqrt{5}}{2}\) from the quadratic formula, students often make errors when rewriting this in a cleaner form. They might incorrectly write \(\mathrm{y} = \frac{\sqrt{5} - 1}{2}\) as \(\mathrm{y} = \frac{1 - \sqrt{5}}{2}\), or conversely write \(\mathrm{y} = \frac{-1 - \sqrt{5}}{2}\) as \(\mathrm{y} = -(1 + \sqrt{5})/2\) instead of the correct \(-\frac{\sqrt{5} + 1}{2}\).

Errors while selecting the answer

1. Forgetting the non-negative constraint on square roots
The most critical error at this stage is accepting both solutions from the quadratic equation without checking validity. Since \(\sqrt{\mathrm{x}}\) represents the principal square root of a real number, it must be non-negative. Students often fail to eliminate \(\mathrm{y} = -\frac{\sqrt{5} + 1}{2} \approx -1.618\), which is negative and therefore invalid. They might randomly choose between the two solutions or even select the negative one.

2. Mismatching the final expression with answer choices
Even when students arrive at the correct value \(\frac{\sqrt{5} - 1}{2}\), they sometimes struggle to recognize this matches answer choice A: \(\frac{1}{2}(\sqrt{5} - 1)\). The notation ½ versus 1/2 can cause confusion, leading them to select an incorrect answer choice that looks similar but is mathematically different.