The sum of squares of two consecutive natural numbers is 244 find the numbers

Text Solution

Answer : The required consecutive even natural numbers are 10 and 12.

Solution : Let the two consecutive even natural numbers be x and x+2. <br> Then `x^(2)+(x+2)^(2)=244` <br> `:.x^(2)+x^(2)+4x+4-244=0` <br> `:.2x^(2)+4x-240=0` <br> `:.x^(2)+2x-120=0" "……` (Dividing by 2) <br> `:.x^(2)+12x-10-120=0` <br> `:.x(x+12)-10(x+12)=0` <br> `:.(x+12)(x-10)=0` <br> `:.x+12=0orx-10=0` <br> `:.x=-12orx=10` <br> But -12 is not a natural number. <br> `:.x=-2` is unacceptable. x=10 and x+2=10+2=12

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The sum of squares of two consecutive natural numbers is 244 ; find the numbers.

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.