Question

if a+b+c= 12 and a^2+b^2+c^2=64,

find the value of :
ab+bc+ac​

Answer 1

Answer:

Step-by-step explanation:

Given a+b+c=12 ,

a^2+b^2+c^2=64 ,

Then ab+b c+ca=?

a+b+c=12 ,

Squaring on both sides ,

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

a^2+b^2+c^2+2(ab+b c+ca)=144 ,

64+2 (ab+b c+ca)=144 ,

2(ab+b c+ca)=80 ,

ab+b c+ca=40 .

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Answer 2

We know a Algebraic Identity :

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

Substitute The Values

⇒ 12² = 64 + 2ab + 2bc + 2ac

⇒ 12² = 64 + 2(ab + bc + ac)

⇒ 144 = 64 + 2(ab + bc + ac)

Subtracting 64 from Both Sides

⇒ 144 - 64 = 64 + 2(ab + bc + ac) - 64

⇒ 80 = 2(ab + bc + ac)

Dividing Both Sides by 2

⇒ 80/2 = [2(ab + bc + ac)]/2

⇒ 40 = ab + bc + ac