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 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.
|