Question

If pq = -6 and 16p2+ 16q2= 208, find the value of (4p + 4q)2?

Answer 1

$\sf \purple{given}$

$\sf pq = - 6$

$\sf 16 {q}^{2} + 16 {p}^{2} = 208$

$\implies \sf {(4q)}^{2} + {(4p)}^{2} = 208$

$\blue{ \boxed{ {a}^{2} + { b}^{2} = {( a+b )}^{2} - 2ab}}$

$\implies \sf {(4q + 4p)}^{2} - 2(4q)(4p)= 208$

$\implies \sf {(4p + 4q)}^{2} - 32pq= 208$

$\implies \sf {(4p + 4q)}^{2} - 32( - 6)= 208$

$\implies \sf {(4p + 4q)}^{2} + 192= 208$

$\implies \sf {(4p + 4q)}^{2} = 208 - 192$

$\orange{\therefore \sf {(4p + 4q)}^{2} = 16}$

HOPE IT HELPS!!