Grade 8 - Algebra #1

Algebra Level 3

Four real numbers w , x , y , z w,~x,~y,~z satisfy w 2 = 7 , x 2 = 3 , y 2 = 5 , z 2 = 6 w^2=7,~x^2=3,~y^2=5,~z^2=6 .

Find the value of:

( − w + x + y + z ) 2 + ( w − x + y + z ) 2 + ( w + x − y + z ) 2 + ( w + x + y − z ) 2 (-w+x+y+z)^2+(w-x+y+z)^2+(w+x-y+z)^2+(w+x+y-z)^2

This problem is a part of <Grade 8 - Algebra> series .


The answer is 84.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jul 8, 2017

I'll use the term "sign" for the "+" and "-" that comes in front of a variable.

For example, the "sign" of − 2 x -2x is "-", and the "sign" of 7.9 y 7.9y is "+", and so on. The actual variable's negativity or positivity doesn't affect.

Choose any two numbers from w , x , y , z w,~x,~y,~z and you realize that two terms have them with different "sign"s, and the others have them with same "sign"s, which, when expanded, will add up to 0 0 .

Therefore, the original expression is equivalent to:

4 ( w 2 + x 2 + y 2 + z 2 ) = 4 × 21 = 84 4(w^2+x^2+y^2+z^2)=4\times21=\boxed{84}

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