Question

if a+b+c =12 and a^2+b^2+c^2=64, find the value of ab+bc+ac

Answer 1

(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca
12^2=64+2(ab+bc+ca)
144-64=2(ab+bc+ca)
80/2=ab+bc+ca
ab+bc+ca=40

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