Question

Difficulty: HardFunction Definitions, Evaluation, and Custom Operators

For all non-zero real numbers xx and yy, the custom operation \star is defined by xy=xyyxx \star y = \frac{x}{y} - \frac{y}{x}. If pp and qq are positive numbers such that p+q=5p + q = 5 and pq=56p \star q = \frac{5}{6}, what is the value of (p+1)q(p + 1) \star q?

  1. 32\frac{3}{2}Answer
  2. B
    32-\frac{3}{2}
  3. C
    245\frac{24}{5}
  4. D
    00
  5. E
    2110\frac{21}{10}

Answer

32\frac{3}{2}
By simplifying pq=p2q2pq=(pq)(p+q)pqp \star q = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}, we substitute p+q=5p + q = 5 to get 5(pq)pq=56\frac{5(p - q)}{pq} = \frac{5}{6}, which reduces to 6(pq)=pq6(p - q) = pq. Substituting q=5pq = 5 - p yields the quadratic equation p2+7p30=0p^2 + 7p - 30 = 0. Since p>0p > 0, we find p=3p = 3 and q=2q = 2. Evaluating (3+1)2=42=4224=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}.

Step-by-Step Solution

1
Express the custom operation using algebraic fractions and factor the numerator
pq=pqqp=p2q2pq=(pq)(p+q)pqp \star q = \frac{p}{q} - \frac{q}{p} = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}
Finding a common denominator converts the expression into a form where known identities can be substituted.
2
Substitute p+q=5p + q = 5 and pq=56p \star q = \frac{5}{6} into the simplified expression
5(pq)pq=56    pqpq=16    6(pq)=pq\frac{5(p - q)}{pq} = \frac{5}{6} \implies \frac{p - q}{pq} = \frac{1}{6} \implies 6(p - q) = pq
Simplifying by dividing both sides by 5 establishes a direct relation between the difference and product of p and q.
3
Substitute q=5pq = 5 - p to form and solve a quadratic equation for pp
6(2p5)=p(5p)    12p30=5pp2    p2+7p30=0    (p+10)(p3)=06(2p - 5) = p(5 - p) \implies 12p - 30 = 5p - p^2 \implies p^2 + 7p - 30 = 0 \implies (p + 10)(p - 3) = 0
Since p and q are positive numbers, p must be positive, giving p = 3 and q = 2.
4
Evaluate (p+1)q(p + 1) \star q with p=3p = 3 and q=2q = 2
(3+1)2=42=4224=212=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = 2 - \frac{1}{2} = \frac{3}{2}
Applying the custom operator definition to inputs 4 and 2 yields the final value.

Key Concept

Custom Operator Evaluation and Quadratic System Solving
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