Question

if a2+b2+c2=49 and ab+bc+ca=36, find a+b+c.​

Answer 1

Answer:

a+b+c = 11

Step-by-step explanation:

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

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

$(a+b+c)^{2}$ = 49+2*36

$(a+b+c)^{2}$ = 121    (then find the square root of 121 that is 11)

so,    

a+b+c = 11

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