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Find three consecutive even integers such that the sum of the squares of the first two is four less than twice the square of the third.

Question

Find three consecutive even integers such that the sum of the squares of the first two is four less than twice the square of the third.

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Step 1

Let the three consecutive even integers are n , n+2, n+4

Step 2

The sum of square of n and sqaure of n+2 is four less than twice the square of n+4

n (n+2)2(n+4)-4
nn4n4 = 2(n' +8n+16-4
2n24n4 2n2 16n+32-4
12n 24 0
12n 24
n -2
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n (n+2)2(n+4)-4 nn4n4 = 2(n' +8n+16-4 2n24n4 2n2 16n+32-4 12n 24 0 12n 24 n -2

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Step 3

Thus the numbers are n=-2. n+2=-2...

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Tagged in

Math

Algebra

Equations and In-equations

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