Question:\(45\mathrm{x}^2 + (15\mathrm{a} + 9\mathrm{b})\mathrm{x} + 3\mathrm{ab} = 0\)In the given equation, a and b are positive constants. The prod...
GMAT Advanced Math : (Adv_Math) Questions
\(45\mathrm{x}^2 + (15\mathrm{a} + 9\mathrm{b})\mathrm{x} + 3\mathrm{ab} = 0\)
In the given equation, \(\mathrm{a}\) and \(\mathrm{b}\) are positive constants. The product of the solutions to the given equation is \(\mathrm{kab}\), where \(\mathrm{k}\) is a constant. What is the value of \(\mathrm{k}\)?
\(\frac{1}{45}\)
\(\frac{1}{15}\)
\(\frac{1}{9}\)
\(\frac{1}{3}\)
1. TRANSLATE the problem information
- Given equation: \(45\mathrm{x}^2 + (15\mathrm{a} + 9\mathrm{b})\mathrm{x} + 3\mathrm{ab} = 0\)
- \(\mathrm{a}\) and \(\mathrm{b}\) are positive constants
- Product of solutions equals \(\mathrm{kab}\), where \(\mathrm{k}\) is unknown
2. INFER the approach needed
- This is asking about the product of roots of a quadratic equation
- For any quadratic in standard form \(\mathrm{Ax}^2 + \mathrm{Bx} + \mathrm{C} = 0\), Vieta's formulas tell us the product of roots equals \(\mathrm{C}/\mathrm{A}\)
- I need to identify \(\mathrm{A}\) and \(\mathrm{C}\), then set up an equation to find \(\mathrm{k}\)
3. TRANSLATE the quadratic into standard form
- Standard form: \(\mathrm{Ax}^2 + \mathrm{Bx} + \mathrm{C} = 0\)
- From \(45\mathrm{x}^2 + (15\mathrm{a} + 9\mathrm{b})\mathrm{x} + 3\mathrm{ab} = 0\):
- \(\mathrm{A} = 45\) (coefficient of \(\mathrm{x}^2\))
- \(\mathrm{B} = (15\mathrm{a} + 9\mathrm{b})\) (coefficient of \(\mathrm{x}\))
- \(\mathrm{C} = 3\mathrm{ab}\) (constant term)
4. INFER the product of solutions using Vieta's formulas
- Product of solutions = \(\mathrm{C}/\mathrm{A} = 3\mathrm{ab}/45\)
5. SIMPLIFY the fraction
- \(3\mathrm{ab}/45 = \mathrm{ab}/15\) (dividing numerator and denominator by 3)
6. TRANSLATE the given condition into an equation
- Problem states: product of solutions = \(\mathrm{kab}\)
- From our work: product of solutions = \(\mathrm{ab}/15\)
- Therefore: \(\mathrm{kab} = \mathrm{ab}/15\)
7. SIMPLIFY to solve for k
- \(\mathrm{kab} = \mathrm{ab}/15\)
- Divide both sides by \(\mathrm{ab}\) (valid since \(\mathrm{a}, \mathrm{b} > 0\)): \(\mathrm{k} = 1/15\)
Answer: (B) 1/15
Why Students Usually Falter on This Problem
Most Common Error Path:
Missing conceptual knowledge: Students don't remember or recognize the need for Vieta's formulas.
Without knowing that the product of roots equals \(\mathrm{C}/\mathrm{A}\), students might try to actually solve the quadratic equation using the quadratic formula, which becomes very messy with the parameters \(\mathrm{a}\) and \(\mathrm{b}\). This leads to getting overwhelmed by complex algebra and ultimately guessing.
Second Most Common Error:
Weak SIMPLIFY execution: Students correctly identify that product = \(3\mathrm{ab}/45\) but make arithmetic errors when simplifying.
Some students might incorrectly simplify \(3\mathrm{ab}/45\), perhaps getting \(1/5\) instead of \(1/15\) by incorrectly reducing the fraction. This could lead them to select Choice (D) (1/3) if they make multiple calculation mistakes.
The Bottom Line:
This problem tests whether students recognize the connection between the abstract statement "product of solutions is \(\mathrm{kab}\)" and the concrete application of Vieta's formulas. The key insight is that you don't need to solve the quadratic explicitly - Vieta's formulas provide a direct shortcut to the product of roots.
\(\frac{1}{45}\)
\(\frac{1}{15}\)
\(\frac{1}{9}\)
\(\frac{1}{3}\)