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rk21857
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Can you explain the part where you take the 4 out of the equation in layman's terms. please

Just factor out 4 out of \(4x^2+4y^2+4z^2\): \((2x)^2+(2y)^2+(2z)^2=4x^2+4y^2+4z^2=4*(x^2+y^2+z^2)\).
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4x^2+4y^2+4z^2 : How do you know which number to manipulate it to. How do you know not to put a 6 or 8 as the coefficient instead of 4, or whether to leave it as 2 or change it to something else.
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4x^2+4y^2+4z^2 : How do you know which number to manipulate it to. How do you know not to put a 6 or 8 as the coefficient instead of 4, or whether to leave it as 2 or change it to something else.

Please read the stem: "If a, b and c are even integers..." So a, b and c can be expressed as a=2x, b=2y, and c=2z for some integers x, y, and z --> a^2+b^2+c^2=(2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2).
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why does 2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2)? If you multiply 4 by each variable wouldnt that make 4x^2+4y^2+4z^2. I read the question stem Im just not getting it....
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why does 2x)^2+(2y)^2+(2z)^2=4(x^2+y^2+z^2)? If you multiply 4 by each variable wouldnt that make 4x^2+4y^2+4z^2. I read the question stem Im just not getting it....

I don't quite understand your question but anyway, step by step:

\(a^2+b^2+c^2=(2x)^2+(2y)^2+(2z)^2=4x^2+4y^2+4z^2=4*(x^2+y^2+z^2)\).
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My answer is A.

there is atleast one two in every even integer,in other words even integer can be expressed as 2 times something

and squares will have atleast one 4 in them.

so answer should be a multiple of 4 , out of the given answers only A is multiple of 4.
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Although above solutions are more elegant than mine, you can also simply list some even perfect squares

\(4,16,36,64,100,144,256,324,400\)

then, through trial and error, \(4+36+100\) produces a match
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a = 2x
b = 2y
c=2z

a^2 + b^2 + c^2 = (2x)^2 + (2y)^2 + (2z)^2
=4(x^2 + y^2 + z^2)
= so the value should be a multiple of 4
-> only A is a multiple of 4
So, Answer is A
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