Question:In physics, the relationship between two variables can sometimes be modeled by equations of the form x + k/x =...
GMAT Advanced Math : (Adv_Math) Questions
In physics, the relationship between two variables can sometimes be modeled by equations of the form \(\mathrm{x + \frac{k}{x} = c}\). The equation \(\mathrm{x + \frac{k}{x} = 6}\) has two solutions. One of the solutions can be written as \(\mathrm{3 + \sqrt{5}}\). What is the value of k?
1. SIMPLIFY the rational equation to quadratic form
- Given: \(\mathrm{x + \frac{k}{x} = 6}\) with solution \(\mathrm{x = 3 + \sqrt{5}}\)
- Multiply both sides by x to eliminate the fraction:
\(\mathrm{x^2 + k = 6x}\) - Rearrange to standard quadratic form:
\(\mathrm{x^2 - 6x + k = 0}\)
2. SIMPLIFY using the quadratic formula
- Apply the quadratic formula with \(\mathrm{a = 1, b = -6, c = k}\):
\(\mathrm{x = \frac{6 ± \sqrt{36 - 4k}}{2}}\) - Factor and simplify the discriminant:
\(\mathrm{x = \frac{6 ± \sqrt{4(9 - k)}}{2}}\)
\(\mathrm{= \frac{6 ± 2\sqrt{9 - k}}{2}}\)
\(\mathrm{= 3 ± \sqrt{9 - k}}\)
3. INFER the relationship using the given solution
- The general solution is: \(\mathrm{x = 3 ± \sqrt{9 - k}}\)
- One specific solution is given as: \(\mathrm{x = 3 + \sqrt{5}}\)
- Match these forms: \(\mathrm{3 + \sqrt{5} = 3 + \sqrt{9 - k}}\)
4. SIMPLIFY to solve for k
- From the equation above: \(\mathrm{\sqrt{5} = \sqrt{9 - k}}\)
- Square both sides: \(\mathrm{5 = 9 - k}\)
- Solve for k: \(\mathrm{k = 9 - 5 = 4}\)
Answer: 4
Why Students Usually Falter on This Problem
Most Common Error Path:
Weak SIMPLIFY skills: Students attempt to substitute \(\mathrm{x = 3 + \sqrt{5}}\) directly into the original equation \(\mathrm{x + \frac{k}{x} = 6}\) without first converting to quadratic form. This creates a complex algebraic mess: \(\mathrm{(3 + \sqrt{5}) + \frac{k}{(3 + \sqrt{5})} = 6}\), which requires rationalizing denominators and becomes unnecessarily complicated. This leads to confusion and guessing.
Second Most Common Error:
Inadequate SIMPLIFY execution: Students correctly get to \(\mathrm{\sqrt{5} = \sqrt{9 - k}}\) but make errors when squaring both sides, either forgetting to square both terms or making computational mistakes like \(\mathrm{5 = 9 - k → k = 5 - 9 = -4}\). This may lead them to enter -4 instead of the correct answer.
The Bottom Line:
This problem tests whether students recognize that working with the general solution form (via the quadratic formula) is more efficient than trying to substitute the specific solution directly. The key insight is that matching solution forms allows for clean algebraic manipulation.