Question

The sum of squares of two consecutive odd positive integers is 202. Find the integers.

Answer 1

$Let \: x \:and \: (x+2) \:are \: two \: consecutive \\odd \: positive \: integers .$

/* According to the problem , given */

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

$\implies x^{2} + x^{2} + 4x + 4 - 202 = 0$

$\implies 2x^{2} + 4x - 198 = 0$

/* On Dividing each term by 2 ,we get */

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

/* Splitting the middle term,we get */

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

$\implies x( x + 11) - 9( x + 11 ) = 0$

$\implies ( x + 11 )( x - 9 ) = 0$

$\implies ( x + 11 ) = 0 \: Or \: ( x - 9 ) = 0$

$\implies x = - 11 ) \: Or \: x = 9$

/* x should not be negative */

Therefore.,

$x = 9$

$\red { Required \: two }$

$\red { positive \:odd \: numbers \: are }\green {= 9,\: 11 }$

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Answer 2

$\huge\bold\green{Question}$

The sum of squares of two consecutive odd positive integers is 202. Find the integers.

$\huge\bold\green{Answer}$

Let us assumed two consecutive odd positive integers

→ x

→ (x + 1)

According to the question , The sum of squares of two consecutive odd positive integers is 202 :-

→ x² + (x+2)² = 202

Now by using identity

[ (a+b)² = a² + b² + 2ab ]

→ x² + x² + 4x + 4 - 202 = 0

→ 2x² + 4x - 198 = 0

Now we divide these equation with 2 and we get :-

→ x² + 2x - 99 = 0

Now by using middle term splitting method :-

→ x² + 11x - 9x - 99 = 0

→ x( x + 11) - 9( x + 11 ) = 0

→ ( x + 11 )( x - 9 ) = 0

→ ( x + 11 ) = 0 and ( x - 9 ) = 0

→ x = - 11 and x = 9

As said in question integers should be positive and odd

Hence the required value of positive odd numbers are 9 and 11