Question

if a^2+b^2+c^2 = 74 and ab+bc+ac = 61 find a+b+c
Plzz answer

Answer 1

Answer:

14

Step-by-step explanation:

We know that

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

Therefore by putting values on the right of equality

(a+b+c)^2= 74+2(61)

= 196

Therefore

a+b+c = √196 = 14 ans

Answer 2

Answer:

Step-by-step explanation:

(a+(b+c))^2

Let A= a

B= (b+c)

So,

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

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

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

(a+(b+c))^2= 74+ 2(61)[since a^2+b^2+c^2= 74 and ab+bc+ca = 61]

(a+(b+c))^2= 196

Therefore,

(a+(b+c))= (196)^1/2 (i.e square root of 196)= 14

Therefore,

a+b+c= 14.

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