Question

If a + b + c = 14 & a² + b² + c² = 50, find ab + bc + ca

Answer 1

EXPLANATION.

⇒ a + b + c = 14.

⇒ a² + b² + c² = 50.

As we know that,

Formula of :

⇒ (x + y + z)² = x² + y² + z² + 2xy + 2yz + 2zx.

⇒ (x + y + z)² = x² + y² + z² + 2(xy + yz + zx).

Using this formula in the equation, we get.

⇒ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca).

Put the values in the equation, we get.

⇒ (14)² = 50 + 2(ab + bc + ca).

⇒ 196 = 50 + 2(ab + bc + ca).

⇒ 196 - 50 = 2(ab + bc + ca).

⇒ 146 = 2(ab + bc + ca).

⇒ 73 = ab + bc + ca.

⇒ ab + bc + ca = 73.

Answer 2

Given:-

• a + b + c = 14

• a² + b² + c² = 50

To find :-

• ab + bc + ca

Solution:-

We know that ,

${\small{\bold{\purple{\underline{(a+b+c)^{2}=a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)}}}}}$

${\sf{\implies{\bf{(14)^{2}=50+2(ab+bc+ca)}}}}$

${\sf{\implies{\bf{196=50+2(ab+bc+ca)}}}}$

${\sf{\implies{\bf{196-50=2(ab+bc+ca)}}}}$

${\sf{\implies{\bf{146=2(ab+bc+ca)}}}}$

${\sf{\implies{\bf{\dfrac{146}{2}=(ab+bc+ca)}}}}$

${\sf{\implies{\bf{73=(ab+bc+ca)}}}}$

${\implies{\small{\bold{\pink{\underline{(ab+bc+ca)=73}}}}}}$

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