Question

if p-q=6 and PQ=10 find (p+q)^2

Answer 1

I'm solving the same question twice cuz uh have asked twice -,-

$\sf \: Solution$

$\sf {(p - q)}^{2} = {p}^{2} + {q}^{2} - 2pq$

$\sf {6}^{2} = {p}^{2} + {q}^{2} - 2(10)$

$\sf \: 36 = {p}^{2} + {q}^{2} - 20$

$\sf \: 36 + 20 = {p}^{2} + {q}^{2}$

$\sf {p}^{2} + {q}^{2} = 56$

Now,

$\bf {(p + q)}^{2} = {p}^{2} + {q}^{2} + 2pq$

$\sf {(p + q)}^{2} = 56 + 20 = \pink{ 76}$

Answer 2

Step-by-step explanation:

p-q=6

pq=10

(p+q)^2=?

(a+b)^2=(a-b)^2+4ab

(p+q)^2=(p-q)^2+4pq

36+4×10

36+40

=76