Question

If $a+b+c=9$ and $ab+bc+ca=40$. Find the value of $a^2+b^2+c^2$.

Answer 1

Given, $a+b+c=9$

and $ab+bc+ca=40$

We know that,
$(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$

$\Rightarrow a^2+b^2+c^2=(a+b+c{)}^2–2(ab+bc+ca)$

$\Rightarrow a^2+b^2+c^2=(9{)}^2-2\times 40=81–80=1$

[$\because a+b+c=9$ and $ab+bc+ca=40$]

Thus, the value of $a^2+b^2+c^2$ is $1.$