Question

if a + b + c = 12 and a2+ b2 + c 2= 64 find the value of ab + bc + ac

Answer 1

As per the identity,
(a + b + c)² = a² + b² + c² + 2.a.b + 2.b.c + 2.c.a

According to question,
a + b + c = 12 and a² + b² + c² = 64.

Using this in the identity,

(12)² = 64 + 2.(a.b + b.c + a.c)
=> 2.(a.b + b.c + a.c) = 144 - 64.
=> 2.(a.b + b.c + a.c) = 80
=> (a.b + b.c + a.c) = 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