Let n be a positive integer such that five times n equals three less than twice the square of n (that is, 5n = 2n^2 − 3). What is the value of n?
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A3
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B13
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C23
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D33
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E6
Answer
Correct Answer: 3
Explanation
Introduction / Context:Here we turn a verbal relation into a quadratic equation in a positive integer. Recognizing and solving the quadratic cleanly, then applying the positivity constraint, identifies the valid integer solution. This is a direct test of forming and solving quadratic equations.
Given Data / Assumptions:
- n is a positive integer.
- Equation: 5n = 2n^2 − 3.
- We need the value of n that satisfies the equation.
Concept / Approach:Rearrange the equation into standard quadratic form an^2 + bn + c = 0, then factor or use the quadratic formula. Discard any non-integer or negative solution that does not meet the stated condition on n.
Step-by-Step Solution:
Move all terms to one side: 2n^2 − 5n − 3 = 0.Compute discriminant: D = (−5)^2 − 4 * 2 * (−3) = 25 + 24 = 49.Square root of D: sqrt(49) = 7.Solutions: n = (5 ± 7) / (2 * 2) = (5 ± 7) / 4.Hence n = 12 / 4 = 3 or n = −2 / 4 = −1/2.Since n is a positive integer, n = 3.Verification / Alternative check:Check: Left side 5n = 15; right side 2n^2 − 3 = 2 * 9 − 3 = 15. The equality holds.
Why Other Options Are Wrong:13, 23, and 33 do not satisfy the equation. 6 does not satisfy it either when substituted.
Common Pitfalls:Sign errors when moving terms, miscomputing the discriminant, or accepting a negative or fractional root despite the integer requirement.
Final Answer:3