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If $a + b + c = 6$ and $a^2 + b^2 + c^2 = 14$, find $ab + bc + ca$.
A$11$
B$13$
C$10$
D$9$
Answer & Solution
Correct answer: A. $11$
**Identity.** $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$.
**Plug in.** $36 = 14 + 2(ab+bc+ca) \;\Rightarrow\; ab+bc+ca = \dfrac{22}{2} = 11$.
**Why option B is tempting.** Subtracting $14$ from $36$ gives $22$, and half-remembered identities lead some to write $22$ or $10$. The factor of two on the cross-product term is the load-bearing detail.
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