If 3p^2 = 5p + 2 and 3q^2 = 5q + 2 where p ne | vedclass

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Question

(A) Given $3p^2 - 5p - 2 = 0$ and $3q^2 - 5q - 2 = 0$,$p$ and $q$ are roots of $3x^2 - 5x - 2 = 0$.Thus,$p + q = \frac{5}{3}$ and $pq = -\frac{2}{3}$.

Vedclass · Accepted Answer

$3x^2 - 5x - 100 = 0$

Vedclass · Answer

(A) Given $3p^2 - 5p - 2 = 0$ and $3q^2 - 5q - 2 = 0$,$p$ and $q$ are roots of $3x^2 - 5x - 2 = 0$.Thus,$p + q = \frac{5}{3}$ and $pq = -\frac{2}{3}$.Let the roots of the new equation be $\alpha = 3p - 2q$ and $\beta = 3q - 2p$.Sum of roots $\alpha + \beta = (3p - 2q) + (3q - 2p) = p + q = \frac{5}{3}$.Product of roots $\alpha \beta = (3p - 2q)(3q - 2p) = 9pq - 6p^2 - 6q^2 + 4pq = 13pq - 6(p^2 + q^2)$.Since $p^2 + q^2 = (p + q)^2 - 2pq = (\frac{5}{3})^2 - 2(-\frac{2}{3}) = \frac{25}{9} + \frac{4}{3} = \frac{37}{9}$.Product $\alpha \beta = 13(-\frac{2}{3}) - 6(\frac{37}{9}) = -\frac{26}{3} - \frac{74}{3} = -\frac{100}{3}$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^2 - (\frac{5}{3})x - \frac{100}{3} = 0 \implies 3x^2 - 5x - 100 = 0$.

If $3p^2 = 5p + 2$ and $3q^2 = 5q + 2$ where $p \neq q$,then the equation whose roots are $3p - 2q$ and $3q - 2p$ is:

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