Question

find the value of p if (3'2)ina solutions of px -5y+p^2​

Answer 1

p = 2 or p = -5.

Step-by-step explanation:

Given :

$px \: \: - \: 5y \: \: + \: {p}^{2}$

its solution = (3, 2) i.e. x = 3 and y = 2

To find :

The value of p.

Solution :

Since, x = 3 and y = 2, then if we put these values in the given expression, value of the expression should be zero.

therefore,

$p \times 3 \: \: - \: 5 \times 2 \: \: + \: {p}^{2} \: = 0$

${p}^{2} + 3p - 10 = 0$

Middle term splitting

${p}^{2} + 5p - 2p - 10 = 0 \\ \implies \: p(p + 5) - 2(p + 5) = 0 \\ \implies \: (p - 2)(p + 5) = 0 \\ \implies \: (p - 2) = 0 \: \: or \: \: ( p+ 5) = 0 \\ \implies \: p \: = 2 \: \: or \: \: p \: = - 5$

Hence, p = 2 or p = -5