Question

Find The value of p.
$\sqrt{10} = \sqrt{(4 - 3) ^{2} + (1 - p)^{2} }$
Urgent, Answer Correctly​

Answer 1

$\huge\underline{\overline{\mid{\bold{\blue{\mathcal{ANSWER}}\mid}}}}$

$\sqrt{10} = \sqrt{(4 - 3) {}^{2} + (1 - p) {}^{2} } \\ \sqrt{10} = \sqrt{(1) {}^{2} + (1 - p) {}^{2} }$

squaring both sides we get

$(\sqrt{10} ) {}^{2} = ( \sqrt{(1) {}^{2} + (1 - p) {}^{2} } ) {}^{2}$

$10 = 1 + (1 - p) {}^{2}$

by using identity (a-b)² = a² + b² + 2ab

$10 = 1 + (1 {}^{2} + p {}^{2} - 2p) \\ 10 = 1 + 1 + p {}^{2} - 2p \\ 10 = 2 + p {}^{2} - 2p \\ p {}^{2} - 2p - 8 = 0 \\ p {}^{2} - 4p +2p - 8 = 0 \\ p(p -4) - 2(p - 4) = 0 \\ (p - 4)(p +2) = 0 \\ p = 4 \\ p = -2$

so, p = 4, -2

Answer 2

Answer:

$\therefore \sqrt{10} = \sqrt{(4 - 3) ^{2} + (1 - p)^{2} }$

$\implies \sqrt{10} = \sqrt{1 + 1 - 2p - p^{2}}$

Squaring both sides,

$\implies 10 = 2 - 2p - {p}^{2}$

$\implies {p}^{2} - 2p + 8 = 0$

$\implies {p}^{2} - 2p + 4p + 2 \times 4 = 0$

$\implies p(p - 2) + 4(p - 2) = 0$

$\implies (p + 4)(p - 2) = 0$

$\therefore p + 4 = 0 \\ \implies \: \bf \rm \blue{ p = - 4 }\\ \therefore \: p - 2 = 0 \\ \implies \bf \rm \: \blue{p = 2}$

Hence, Value of p is -4 or 2.