Question

If a²+b²+c²=16 and ab+bc+ca=10, find the value of a+b+c.

Answer 1

$\bold\red{\underline{\underline{Answer:}}}$

$\bold{The \ value \ of \ a+b+c \ is \ 6.}$

$\bold\blue{Explanation}$

$\bold{(a+b+c)^{2}=(a+b+c)(a+b+c)}$

$\bold{=>a^{2}+ab+ac+ab+b^{2}+bc+ac+bc+c^{2}}$

$\bold{=>a^{2}+b^{2}+c^{2}+2(ab+bc+ac)}$

$\bold\orange{Given:}$

$\bold{=>a^{2}+b^{2}+c^{2}=16}$

$\bold{=>ab+bc+ca=10}$

$\bold\pink{To \ find:}$

$\bold{The \ value \ of \ a+b+c}$

$\bold\green{\underline{\underline{Solution}}}$

$\bold{=>a^{2}+b^{2}+c^{2}=16...(1)}$

___________________________________

$\bold{=>ab+bc+ca=10}$

$\bold{Multiply \ both \ sides \ by \ 2}$

$\bold{,we \ get}$

$\bold{=>2(ab+bc+ca)=20...(2)}$

___________________________________

$\bold{From \ (1) \ and \ (2),}$

$\bold{According \ to \ identity,}$

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

$\bold{(a+b+c)^{2}=16+20}$

$\bold{(a+b+c)^{2}=36}$

$\bold{On \ taking \ square \ root \ of \ both \ sides,}$

$\bold{we \ get}$

$\bold{=>a+b+c=6}$

$\bold\purple{\tt{\therefore{The \ value \ of \ a+b+c \ is \ 6.}}}$