Question

if
${a}^{2} + b {}^{2} + c { }^{2} = 20$


and a + b + C
$= 0$
find a b + bc+ ca​

Answer 1

Answer:

$- 10$

Step-by-step explanation:

${a}^{2} + {b }^{2} + {c}^{2} = {(a + b + c)}^{2} - 2(ab + bc + ca) \\ = 20 = {0}^{2} - 2(ab + bc + ca) \\ = 2(ab + bc + ca) = - 20 \\ = ab + bc + ca = \frac{ - 20}{2} \\ = ab + bc + ca = - 10$

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Answer 2

Answer :-

ab + bc + ac = - 10

Explanation :-

Given :-

• a² + b² + c² = 20

• a + b + c =0

To find :-

ab + bc + ca

Solution :-

We know that :-

(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac

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

Here, • a² + b² + c² = 20

• a + b + c =0

By substituting the values

⇒ (0)² = (20) + 2(ab + bc + ac)

⇒ 0 = 20 + 2(ab + bc + ac)

⇒ 0 - 20 = 2(ab + bc + ac)

⇒ - 20 = 2(ab + bc + ac)

⇒ - 20/2 = ab + bc + ac

⇒ - 10 = ab + bc + ac

⇒ ab + bc + ac = - 10

Therefore ab + bc + ac = - 10

Verification :-

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

⇒ (0)² = (20) + 2(-10)

⇒ 0 = 20 + 2(-10)

⇒ 0 = 20 - 20

⇒ 0 = 0

Identity used :-

• (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac