Question
If $a^2+b^2+c^2=250$ and $a b+b c+c a=3$, find $a+b+c$.
✓
Answer
Recall the formula
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$a^2+b^2+c^2=250, a b+b c+c a=3$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(a+b+c)^2=250+2 \cdot(3)$
$(a+b+c)^2=256$
$(a+b+c)= \pm 16$
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$a^2+b^2+c^2=250, a b+b c+c a=3$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(a+b+c)^2=250+2 \cdot(3)$
$(a+b+c)^2=256$
$(a+b+c)= \pm 16$
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