If $a + b + c = m$ and $a^{2} + b^{2} + c^{2} = n$ ,
then $2 (a b + b c + c a)$ is _______.

2 m - n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

2 m + n

No worries! We‘ve got your back. Try BYJU‘S free classes today!

m^{2} - n

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

n^{2} - m

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $m^{2} - n$
Given, $a + b + c = m$ and $a^{2} + b^{2} + c^{2} = n$ .

Using the identity $(a + b + c)^{2}$ = $a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$ ,

$\Rightarrow$ $m^{2} = n + 2 (a b + b c + c a)$

$\Rightarrow$ $2 (a b + b c + c a) = m^{2} - n$ .

Suggest Corrections

Similar questions

a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) :

\frac{(a + b + c)^{2} - a^{2} - b^{2} - c^{2}}{2 (a b + b c + c a)}

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

a^{2} + b^{2} + 2 (a b + b c + c a)

(a + b + m)

(a + b + n c)

(m + n)

Join BYJU'S Learning Program