Find two consecutive odd natural numbers, the sum of whose squares is 202.

We have,

Let the two odd consecutive numbers

So,

According to given question,

If,

If,

So,

Hence, this is the answer.

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find two consecutive odd natural numbers the sum of whose square is 394

Solution

Let the two consecutive odd numbers be x and x+2. ATQ, X2 + (X+2)2 = 394 X2 + X2 + 4X + 22 = 394 2X2 + 4X + 4 = 394 .......( dividing the equation by 2) X2 +2X =(394 - 4)/2 X2 + 2X = 195 X2 +2X - 195 = 0 X2 + 15X - 13X -195 = 0 .............( by factorization meathod ) X( X + 15 ) - 13( X + 15 ) = 0 ( X + 15 ) ( X - 13 ) X = -15 OR X = 13 CASE 1 If X = -15, then, X + 2 = - 13 CASE 2 If X = 13, then,

X + 2 = 15

Find two consecutive odd natural numbers, the sum of whose squares is $$202$$. 

We have,

Let the two odd consecutive numbers $$x\,\text{and}\,x+2$$

So,

According to given question,

$$ {{x}^{2}}+{{\left( x+2 \right)}^{2}}=202 $$

$$ \Rightarrow {{x}^{2}}+{{x}^{2}}+4+4x=202 $$

$$ \Rightarrow 2{{x}^{2}}+4x=198 $$

$$ \Rightarrow {{x}^{2}}+2x=99 $$

$$ \Rightarrow {{x}^{2}}+2x-99=0 $$

$$ \Rightarrow {{x}^{2}}+\left( 11-9 \right)x-99=0 $$

$$ \Rightarrow {{x}^{2}}+11x-9x-99=0 $$

$$ \Rightarrow x\left( x+11 \right)-9\left( x+11 \right)=0 $$

$$ \Rightarrow \left( x+11 \right)x-9=0 $$

If,

$$ x+11=0 $$

$$ \Rightarrow x=-11 $$

If,

$$ x-9=0 $$

$$ x=9 $$

So,

$$x+2=9+2=11$$

Hence, this is the answer.

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