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Algebra Level 3

{ w + 4 x + 9 y + 16 z = 1 4 w + 9 x + 16 y + 25 z = 12 9 w + 16 x + 25 y + 36 z = 123 \begin{cases} w+4x+9y+16z=1\\ 4w+9x+16y+25z=12\\ 9w+16x+25y+36z=123\end{cases}

Let w , x , y , z w,x,y,z be real numbers satisfying the system of equations above.

Find 16 w + 25 x + 36 y + 49 z 16w+25x+36y+49z .


The answer is 334.

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Steve Iroghama by Steve Iroghama , 17, Nigeria

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2 solutions

If we see from the pattern at the problem we can make : a n 2 + b ( n + 1 ) 2 + c ( n + 2 ) 2 = ( n + 3 ) 2 an^2+b(n+1)^2+c(n+2)^2=(n+3)^2 solved it we get : a=1, b=-3, c=3. Where a is the first line, b is the second, and c is the third line. Then the answer become 1 × 1 + 12 × ( − 3 ) + 123 × 3 = 334 1 \times 1 +12 \times (-3) +123 \times 3 = 334

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Steve Iroghama
Feb 3, 2017

Given that

w+4x+9y+16z=1

4w+9x+16y+25z=12

9w+16x+25y+36z=123

To get 16w+25x+36y+49z

=(w+4x+9y+16z) -3(4w+9x+16y+25z) +3(9w+16x+25y+36z)

=1-3(12)+3(123)

=1 - 36 +369

334

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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