Let the First even natural numbers be `x `
Therefore, consecutive even natural numbers be `x + 2`
Sum of squares of these two consecutive even natural numbers is 244
\[x^2 + \left( x + 2 \right)^2 = 244\]
\[ \Rightarrow x^2 + x^2 + 4 + 4x = 244\]
\[ \Rightarrow 2 x^2 + 4x - 240 = 0\]
\[ \Rightarrow x^2 + 2x - 120 = 0\] ....(Dividing both sides by 2)
\[ \Rightarrow x = \frac{- b \pm \sqrt{b^2 - 4ac}}{2a}\]
\[ \Rightarrow x = \frac{- 2 \pm \sqrt{2^2 - 4 \times 1 \times \left( - 120 \right)}}{2}\]
\[ \Rightarrow x = \frac{- 2 \pm \sqrt{4 + 480}}{2}\]
\[\Rightarrow x = \frac{- 2 \pm \sqrt{484}}{2}\]
\[ \Rightarrow x = \frac{- 2 \pm 22}{2}\]
\[ \Rightarrow x = \frac{- 2 + 22}{2}, \frac{- 2 - 22}{2}\]
\[ \Rightarrow x = \frac{20}{2}, \frac{- 24}{2}\]
\[ \Rightarrow x = 10, - 12\]
But the natural number cannot be negative so,
The two numbers are 10 and 10 + 2 = 12.
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The sum of the squares of two consecutive even integers is 724. Find the integers.
Let the first be x. Then the other is x+2.
Have x^2 + (x+2)^2 = 724
So x^2 + (x^2 + 4x + 4) = 724
2x^2 + 4x - 720 = 0
x^2 + 2x - 360 = 0
(x - 18)(x + 20) = 0
So x = 18 since x > 0.
One more line: the two integers are 18 and 20.
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