If a + b + c = 6 and $a^2 + b^2 + c^2 = 26$, then what is ab + bc + ca equal to ?

Question

Accepted Answer

To solve the problem and justify why Option 4 is correct, we need to use the given equations and some algebraic identities.

Given:
1. $ a + b + c = 6 $
2. $ a^2 + b^2 + c^2 = 26 $

We need to find the value of $ ab + bc + ca $.

### Explanation:

We can use the identity:
$$ (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) $$

Substitute the given values into this identity:
$$ 6^2 = 26 + 2(ab + bc + ca) $$

Calculate $ 6^2 $:
$$ 36 = 26 + 2(ab + bc + ca) $$

Now, solve for $ 2(ab + bc + ca) $:
$$ 36 - 26 = 2(ab + bc + ca) $$
$$ 10 = 2(ab + bc + ca) $$

Divide both sides by 2 to find $ ab + bc + ca $:
$$ ab + bc + ca = \frac{10}{2} = 5 $$

Thus, the value of $ ab + bc + ca $ is 5, which corresponds to Option 4.

### Why Other Options are Incorrect:

- **Option 1 (0):** If $ ab + bc + ca $ were 0, then substituting back into the identity would not satisfy the equation $ 36 = 26 + 0 $, which is incorrect.
  
- **Option 2 (4):** If $ ab + bc + ca $ were 4, substituting back would give $ 36 = 26 + 8 $, which simplifies to $ 36 = 34 $, an incorrect statement.

- **Option 3 (2):** If $ ab + bc + ca $ were 2, substituting back would give $ 36 = 26 + 4 $, which simplifies to $ 36 = 30 $, another incorrect statement.

Therefore, the only value that satisfies the equation is 5, making Option 4 the correct answer.