Question

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

Answer 1

Answer:

40 = ab+bc+ac

Step-by-step explanation:

Given,

a + b + c = 12 , a²+b²+c²=64

We know that,

(a + b + c)²= a²+b²+c²+ 2(ab+bc+ac)

(12)²= 64 + 2 ( ab+bc+ac)

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

80/2 = ab+bc+ac

40= ab+bc+ac

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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