Question:\(4\mathrm{x}^2 - (3\mathrm{a} + 2\mathrm{b})\mathrm{x} + \frac{\mathrm{ab}}{2} = 0\)In the given equation, a and b are positive constants. T...
GMAT Advanced Math : (Adv_Math) Questions
\(4\mathrm{x}^2 - (3\mathrm{a} + 2\mathrm{b})\mathrm{x} + \frac{\mathrm{ab}}{2} = 0\)
In the given equation, a and b are positive constants. The product of the solutions to the given equation is \(\mathrm{k} \cdot \mathrm{ab}\), where k is a constant. What is the value of k?
1. TRANSLATE the problem information
- Given equation: \(\mathrm{4x^2 - (3a + 2b)x + ab/2 = 0}\)
- We need to find k where (product of solutions) = \(\mathrm{k \cdot ab}\)
- Constants a and b are positive
2. INFER the solution approach
- Since we need the product of solutions for a quadratic, Vieta's formulas are the most direct method
- For any quadratic \(\mathrm{Ax^2 + Bx + C = 0}\), the product of solutions equals \(\mathrm{C/A}\)
- We need to identify coefficients A and C from our equation
3. TRANSLATE coefficients from standard form
Comparing \(\mathrm{4x^2 - (3a + 2b)x + ab/2 = 0}\) to \(\mathrm{Ax^2 + Bx + C = 0}\):
- \(\mathrm{A = 4}\)
- \(\mathrm{B = -(3a + 2b)}\)
- \(\mathrm{C = ab/2}\)
4. SIMPLIFY to find the product of solutions
- Product = \(\mathrm{C/A = (ab/2)/4 = ab/8}\)
5. SIMPLIFY to solve for k
- We know: product of solutions = \(\mathrm{k \cdot ab}\)
- So: \(\mathrm{ab/8 = k \cdot ab}\)
- Divide both sides by ab: \(\mathrm{k = 1/8}\)
Answer: B) 1/8
Why Students Usually Falter on This Problem
Most Common Error Path:
Missing conceptual knowledge: Not remembering Vieta's formulas or confusing the product formula with the sum formula.
Students might try to actually solve the quadratic equation using the quadratic formula, which becomes unnecessarily complex with the parameters a and b. They get lost in algebraic manipulation and either make calculation errors or abandon the systematic approach entirely. This leads to confusion and guessing.
Second Most Common Error:
Weak TRANSLATE skill: Incorrectly identifying the constant term C in the equation.
Students might misread the constant term as ab instead of \(\mathrm{ab/2}\), leading them to calculate the product as \(\mathrm{(ab)/4 = ab/4}\) instead of \(\mathrm{(ab/2)/4 = ab/8}\). When they set up \(\mathrm{k \cdot ab = ab/4}\), they get \(\mathrm{k = 1/4}\) and select Choice C (1/4).
The Bottom Line:
This problem tests whether students recognize that Vieta's formulas provide an elegant shortcut. The key insight is that we don't need to solve for the actual roots—we just need their product, which Vieta's formulas give us directly.