Question

please give answer as soon as I will mark your answer as brainleist and please please please EXPLAIN​

Answer 1

$\tt \orange {Question}$

$\tt \: If \: x-3 \: and \: x-1/3 \: are \: both \: factors of \: \\ \tt px^2 +5x +r . \: then \: show \: \: \: \: \: \: \: \: p=r$

$\tt \orange {Solution}$

$\tt \: let \: p(x) = {px}^{2} + 5 + r$

Zero of x-1/3 = 3

$\tt \: so$

$\tt \: p(3 {) }^{2} + 5(3) + r = 0$

so 9 p+ r = - 9 ------》 Equation 1.

$\tt \: p( \frac{1}{3} ) {}^{2} + 5( \frac{1}{3} {)}^{2} + r = 0$

$\tt \frac{1}{9} p \: + \frac{5}{9} + r = 0$

$\tt \: p( \frac{1}{9} ) + 1 + r = 0$

$\tt \: p( \frac{1}{9} ) + r = - 1$

$\tt \: p + 9r = -9 ------》 Equation 2.$

From eqn 1 and 2

we get

$\tt \: (9p + r = - 9 ) - (p + 9r = - 9)$

$\tt \: 8p - 8r = 0 \\ \\ \not 8p = \not8r$

${\huge {\bold {\boxed{p = r}}}}$