square letters

Algebra Level 3

Let a , b , c a,b,c real numbers such that

a 2 + 2 b = 7 a^2+2b=7

b 2 + 4 c = − 7 b^2+4c=-7

c 2 + 6 a = − 14 c^2+6a=-14

Find a 2 + b 2 + c 2 a^2+b^2+c^2


The answer is 14.

Paola Ramírez by Paola Ramírez , 24, Mexico

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1 solution

Paola Ramírez
Mar 9, 2015

a 2 + 2 b + b 2 + 4 c + c 2 + 6 a = − 14 a^2+2b+b^2+4c+c^2+6a=-14

Complete the square and get

( a + 3 ) 2 + ( b + 1 ) 2 + ( c + 2 ) 2 = − 14 + 9 + 4 = 0 (a+3)^2+(b+1)^2+(c+2)^2=-14+9+4=0

As any n n number different from zero is bigger than 0 0 if we do n 2 n^2 ⇒ ( a + 3 ) , ( b + 1 ) \Rightarrow(a+3),(b+1) and ( c + 2 ) (c+2) must be equal to 0 0

∴ \therefore

a + 3 = 0 ⇒ a = − 3 a+3=0 \Rightarrow a=-3

b + 1 = 0 ⇒ b = − 1 b+1=0 \Rightarrow b=-1

c + 2 = 0 ⇒ c = − 2 c+2=0 \Rightarrow c=-2

So a 2 + b 2 + c 2 = ( − 3 ) 2 + ( − 1 ) 2 + ( − 2 ) 2 = 14 a^2+b^2+c^2=(-3)^2+(-1)^2+(-2)^2=\boxed{14}

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Note: You should verify that a = − 3 , b = − 1 , c = − 2 a = -3, b = -1, c = -2 is indeed a solution to the original system of linear equations. Because you have manipulated them along the way, you do not have if and only if statements.

Calvin Lin Staff - 6 years, 3 months ago

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