The algebraic expression for the sum of the squares of two consecutive even numbers is:
n^2 + (n+2)^2.
We want to find the value of n for which the sum of the squares is 164:
n^2 + (n+2)^2 = 164.
Expanding the expression:
n^2 + n^2 + 4n + 4 = 164.
Combining like terms:
2n^2 + 4n + 4 = 164.
Subtracting 164 from both sides:
2n^2 + 4n - 160 = 0.
Dividing both sides by 2:
n^2 + 2n - 80 = 0.
Factoring the quadratic equation:
(n + 10)(n - 8) = 0.
Setting each factor equal to zero and solving for n:
n + 10 = 0 -> n = -10,
n - 8 = 0 -> n = 8.
Since the numbers are consecutive even numbers, n cannot be -10. Therefore, the value of n is 8.
Thus, the sum of the squares of the two numbers is:
8^2 + (8+2)^2 = 64 + 100 = 164.
So, the sum of the squares is 164.
Two consecutive even numbers are n and n+2 write down an algebraic for the sum of their squares. Find the number of the sum of the squares is 164
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