Quadratic EquationHard

Question

Let $p,q$ are integers and the two roots of the equation $x^{2} - \left( \frac{p^{2} + 11}{9} \right)x + \frac{15}{4}(p + q) + 16 = 0$ are p and Then

Options

A.$p = 4$
B.$p = 13$
C.$q = 13$
D.$q = 7$

Solution

$p + q = \frac{p^{2} + 11}{9}$

$$\begin{array}{r} \begin{matrix} pq & = \frac{15}{4}(p + q) + 16 & & \\ 4pq - 15(p + q) & = 64 & & \\ (4q - 15)\left( p - \frac{15}{4} \right) & = 64 + \frac{(15)^{2}}{4} = \frac{16^{2} + 15^{2}}{4} & & \\ \Rightarrow \ (4q - 15)(4p - 15) & = 481 = 13 \times 37 & & \\ 4p - 15 = 13, & 4q - 15 & & = 37 \\ = 37 & & & = 13 \\ = 13 \times 37 & & & = 1 \\ = 1 & & \Rightarrow & (p,q) \\ & & & \begin{matrix} 4p & & & = (7,13) \\ & & = & (13,7) \end{matrix} \\ & & & = (124,4) \\ & & & = (4,124) \end{matrix}\#(2) \end{array}$$

Only (p, q) = (13, 7) satisfy (1)

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