Question

The Sum of squares of 2 consecutive natural even numbers is 244; find the numbers.

Answer 1

Let us consider the first even natural number as x, then the second even natural number will be x + 2

It is given that the sum of squares of two consecutive even natural numbers is 244.

This can be written in equation as under:

Then x² + (x + 2)² = 244 × 2 + (x + 2)² = 244

∴ x² + x² + 4x + 4 - 244 = 0

∴ 2x² + 4x - 240 = 0

∴ x² + 2x - 120 = 0      …(Dividing both sides by 2)

∴ x² + 12x − 10 − 120 = 0

∴ x(x + 12) − 10(x + 12) = 0

∴ (x + 12) (x − 10) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 12 = 0 or x − 10 = 0

∴ x = −12 or x = 10

But -12 is not a natural number.

∴ x = −12 is unacceptable.

x = 10 and x + 2 = 10 + 2 = 12

Thus, non be x = 12 & x = 10.

Answer 2

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 

x² + x + 2² = 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, 

x = 10 and x + 2 = 10 + 2 = 12

Thus, non be x = 12 & x = 10.