Question

a+b+c=12 and ab+bc+ca=22 find a2+b2+c2

Answer 1

GIVEN :

a+b+c=12

ab+bc+ca=22

a^2+b^2+c^2=?

We know that

(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ba)

(12)^2=a^2+b^2+c^2+2(22)

144=a^2+b^2+c^2+44

144-44=a^2+b^2+c^2

100=a^2+b^2+c^2

Answer 2

Answer:

$a^2+b^2+c^2=100$

Step-by-step explanation:

Given the equations $a+b+c=12$ and $ab+bc+ca=22$

To solve $(a+b+c)^2$

consider the algebraic identity,

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ba)$

Substituting the given values in the above identity

$(12)^2=a^2+b^2+c^2+2(22)$

Opening the brackets,

$\implies 144=a^2+b^2+c^2+44$

Rearranging the terms, to solve for $a^2+b^2+c^2$,

$\implies a^2+b^2+c^2=144-44$

$\implies a^2+b^2+c^2=100$

Therefore, $a^2+b^2+c^2=100$.