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Question

# If a2+2b=7,b2+4c=−7andc2+6a=−14 then the value of (a2+b2+c2) is

A
14
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B
25
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C
36
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D
47
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Solution

## The correct option is C 14Consider the given equations a2+2b=7, b2+4c=−7 and c2+6a=−14 and add them as follows:a2+2b+b2+4c+c2+6a=7−7−14⇒a2+b2+c2+6a+2b+4c=−14⇒a2+b2+c2+6a+2b+4c+14=0⇒(a2+6a+9)+(b2+2b+1)+(c2+4c+4)=0⇒(a+3)2+(b+1)2+(c+2)2=0(∵(a+b)2=a2+b2+2ab)⇒(a+3)2=0,(b+1)2=0,(c+2)2=0⇒a+3=0,b+1=0,c+2=0⇒a=−3,b=−1,c=−2⇒a2=(−3)2,b2=(−1)2,c2=(−2)2⇒a2=9,b2=1,c2=4Therefore, a2+b2+c2=9+1+4=14Hence, a2+b2+c2=14.

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