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Which of the following equations has a root in common with \(\mathrm{x}^2 - 6\mathrm{x} + 5 = 0\)?
Let's start by understanding what we're actually looking for. The problem asks us to find which equation has "a root in common" with \(\mathrm{x^2 - 6x + 5 = 0}\).
In everyday terms, a "root" is simply a number that, when we substitute it for x, makes the equation equal to zero. Think of it like finding the right key that unlocks a lock - the root is the value that "unlocks" the equation by making it true.
So our task is to:
- First, find what numbers make \(\mathrm{x^2 - 6x + 5 = 0}\) true
- Then check which of the five answer choices is also made true by at least one of those same numbers
Process Skill: TRANSLATE - Converting the mathematical language "root in common" to the concrete task of finding shared solutions
Now let's find the numbers that make \(\mathrm{x^2 - 6x + 5 = 0}\) true.
We need to think: what two numbers multiply to give us 5 and add to give us 6? Since we're looking at \(\mathrm{x^2 - 6x + 5}\), we want factors of 5 that add to 6.
The factors of 5 are: 1 and 5
Let's check: \(\mathrm{1 + 5 = 6}\) ✓ and \(\mathrm{1 × 5 = 5}\) ✓
So we can write: \(\mathrm{x^2 - 6x + 5 = (x - 1)(x - 5) = 0}\)
This tells us that either \(\mathrm{(x - 1) = 0}\) or \(\mathrm{(x - 5) = 0}\)
Therefore: \(\mathrm{x = 1}\) or \(\mathrm{x = 5}\)
So our two roots are \(\mathrm{x = 1}\) and \(\mathrm{x = 5}\). These are the numbers we need to look for in the answer choices.
Now we'll test whether \(\mathrm{x = 1}\) or \(\mathrm{x = 5}\) satisfies any of the given equations. We just need to substitute these values and see if we get zero.
Choice A: \(\mathrm{x^2 + 1 = 0}\)
- Testing \(\mathrm{x = 1}\): \(\mathrm{(1)^2 + 1 = 1 + 1 = 2 ≠ 0}\)
- Testing \(\mathrm{x = 5}\): \(\mathrm{(5)^2 + 1 = 25 + 1 = 26 ≠ 0}\)
Neither root works.
Choice B: \(\mathrm{x^2 - x - 2 = 0}\)
- Testing \(\mathrm{x = 1}\): \(\mathrm{(1)^2 - (1) - 2 = 1 - 1 - 2 = -2 ≠ 0}\)
- Testing \(\mathrm{x = 5}\): \(\mathrm{(5)^2 - (5) - 2 = 25 - 5 - 2 = 18 ≠ 0}\)
Neither root works.
Choice C: \(\mathrm{2x^2 - 2 = 0}\)
- Testing \(\mathrm{x = 1}\): \(\mathrm{2(1)^2 - 2 = 2(1) - 2 = 2 - 2 = 0}\) ✓
We found a match! \(\mathrm{x = 1}\) makes this equation true.
- Testing \(\mathrm{x = 5}\): \(\mathrm{2(5)^2 - 2 = 2(25) - 2 = 50 - 2 = 48 ≠ 0}\)
\(\mathrm{x = 1}\) works, which is enough for a "common root."
Choice D: \(\mathrm{x^2 - 2x - 3 = 0}\)
- Testing \(\mathrm{x = 1}\): \(\mathrm{(1)^2 - 2(1) - 3 = 1 - 2 - 3 = -4 ≠ 0}\)
- Testing \(\mathrm{x = 5}\): \(\mathrm{(5)^2 - 2(5) - 3 = 25 - 10 - 3 = 12 ≠ 0}\)
Neither root works.
Choice E: \(\mathrm{x^2 - 10x - 5 = 0}\)
- Testing \(\mathrm{x = 1}\): \(\mathrm{(1)^2 - 10(1) - 5 = 1 - 10 - 5 = -14 ≠ 0}\)
- Testing \(\mathrm{x = 5}\): \(\mathrm{(5)^2 - 10(5) - 5 = 25 - 50 - 5 = -30 ≠ 0}\)
Neither root works.
Only Choice C: \(\mathrm{2x^2 - 2 = 0}\) shares a root with the original equation \(\mathrm{x^2 - 6x + 5 = 0}\).
Both equations are satisfied when \(\mathrm{x = 1}\), so they have the root \(\mathrm{x = 1}\) in common.
To double-check our work: we can also solve \(\mathrm{2x^2 - 2 = 0}\) directly:
\(\mathrm{2x^2 = 2}\)
\(\mathrm{x^2 = 1}\)
\(\mathrm{x = ±1}\)
So the roots of \(\mathrm{2x^2 - 2 = 0}\) are \(\mathrm{x = 1}\) and \(\mathrm{x = -1}\). Since \(\mathrm{x = 1}\) is also a root of our original equation, we've confirmed they share this common root.
The answer is Choice C: \(\mathrm{2x^2 - 2 = 0}\). This equation shares the root \(\mathrm{x = 1}\) with the original equation \(\mathrm{x^2 - 6x + 5 = 0}\).
1. Misinterpreting "root in common"
Students may think they need to find an equation that has ALL the same roots as the original equation, rather than understanding that "a root in common" means just ONE shared root is sufficient. This leads them to look for equations with both \(\mathrm{x = 1}\) AND \(\mathrm{x = 5}\) as roots, making the problem unnecessarily complex.
2. Planning to solve all answer choice equations completely
Students might plan to find all roots of each answer choice equation first, then compare the complete sets of roots. This is inefficient and time-consuming. The strategic approach is to find the roots of the given equation first, then simply test whether these specific values satisfy each answer choice.
3. Forgetting to find the roots of the original equation first
Some students jump directly into testing answer choices without first determining what the roots of \(\mathrm{x^2 - 6x + 5 = 0}\) actually are. Without knowing that the roots are \(\mathrm{x = 1}\) and \(\mathrm{x = 5}\), they cannot systematically check for common roots.
1. Factoring errors when finding roots of \(\mathrm{x^2 - 6x + 5 = 0}\)
Students may incorrectly factor this as \(\mathrm{(x - 2)(x - 3)}\) or make other factoring mistakes. The correct factoring requires finding two numbers that multiply to +5 and add to +6, which are 1 and 5, giving \(\mathrm{(x - 1)(x - 5) = 0}\).
2. Arithmetic mistakes during substitution
When testing \(\mathrm{x = 1}\) or \(\mathrm{x = 5}\) in the answer choices, students may make calculation errors. For example, when testing \(\mathrm{x = 5}\) in choice C: \(\mathrm{2(5)^2 - 2 = 2(25) - 2 = 50 - 2 = 48}\), students might incorrectly calculate this as 0 due to rushed arithmetic.
3. Sign errors during substitution
Students may make sign mistakes when substituting negative values or when dealing with subtraction. For instance, in choice B when testing \(\mathrm{x = 1}\): \(\mathrm{x^2 - x - 2}\) becomes \(\mathrm{1 - 1 - 2 = -2}\), but students might incorrectly get +2 by mishandling the negative signs.
1. Continuing to check remaining choices after finding the correct answer
On the GMAT, once students find that choice C works (\(\mathrm{x = 1}\) satisfies \(\mathrm{2x^2 - 2 = 0}\)), they should select it and move on. However, some students waste valuable time checking choices D and E even after confirming C is correct, not realizing that only one answer choice will be correct.
2. Selecting the first equation that shares ANY characteristic
Students might select an answer choice because it "looks similar" to the original equation or shares some superficial feature (like being quadratic), without actually verifying that it shares a root. This leads to hasty selection without proper verification through substitution.