Question:
If \[ a^2+b^2+c^2=29 \] and \[ ab+bc+ca=26, \] then \[ a+b+c= \]
(a) 9
(b) 6
(c) 7
(d) 10
Solution:
\[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \]
\[ =(29)+2(26) \]
\[ =29+52 \]
\[ =81 \]
\[ a+b+c=\sqrt{81} \]
\[ =9 \]
\[ \boxed{9} \]
If \[ a^2+b^2+c^2=29 \] and \[ ab+bc+ca=26, \] then \[ a+b+c= \]
(a) 9
(b) 6
(c) 7
(d) 10
\[ (a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca) \]
\[ =(29)+2(26) \]
\[ =29+52 \]
\[ =81 \]
\[ a+b+c=\sqrt{81} \]
\[ =9 \]
\[ \boxed{9} \]