Question

a+b+c=12 and a2+b2+c2=64 find the value of ab+bc+ac

Answer 1

Answer:

Step-by-step explanation:

(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca (formula)

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

12^2= 64+2(ab+bc+ca) ( Substituting the given values)

144= 64+2(ab+bc+ ca)

144–64 = 2(ab+bc+ca) (solving)

80= 2(ab+bc+ ca)

80/2 = ab+bc+ca

40 = ab+ bc+ ca

Ans: 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