Question

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

Answer 1

Answer: A/Q, the given values are

a+b+c = 12

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

Now, the value we need to find is ab + bc + ca. This can be determined by using the expansion of (a + b + c)^2

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

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

(ab+bc+ca) = [(a+b+c)^2 - (a^2+b^2+c^2)]/2

= [(12^2) - (64)]/2

= (144 - 64)/2

= 80/2

ab + bc + ca = 40

Therefore,

ab + bc + ca = 40

Step-by-step explanation:

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