If $p = 0$ $q = 2$ find $p^{2} + 2 p q + q^{2}$ .

Open in App

Solution

if $p = 0$ and $q = 2$
$p^{2} + 2 p q + q^{2} = {(p + q)}^{2}$
$= {(0 + 2)}^{2} = (2^{2}) = 4$
$∴$ $p^{2} + 2 p q + q^{2} = 4$ when $p = 0$ and $q = 2$

Suggest Corrections

Similar questions

(p^{2} - q^{2}) = 2 (p q - q^{2})

P

Q

P Q = 0 \Rightarrow 1) P = 0

Q = 0

b o t h

P Q = I_{2} \Rightarrow P = Q^{- 1}

(P + Q)^{2} = P^{2} + 2 P Q + Q^{2}

p^{2} + q^{2} - a^{2} - b^{2} + 2 p q + 2 a b + p + q - a + b

p^{2} - 2 p q + q^{2}

p^{2} + 2 p q - q^{2}

p = x - y, q = y - z

r = z - r

r^{2} - p^{2} + 2 p q - q^{2}

Join BYJU'S Learning Program