Question

find two consecutive odd positive integer sum of whose square is 290​

Answer 1

Let the two consecutive odd positive integers be x and x + 2

Now it is given that the sum of the squares is 290.

⇒ x² + ( x + 2 )² = 290

⇒ x² + x² + 4x + 4 = 290

⇒ 2x² + 4x + 4 – 290 = 0

⇒ 2x² + 4x – 286 = 0

Dividing by 2 we get,

⇒ x² + 2x – 143 = 0

⇒ x² + 13x – 11x – 143 = 0

⇒ x ( x + 13 ) -11 ( x + 13 ) = 0

⇒ ( x – 11 ) ( x + 13 ) = 0

⇒ x = -13, 11

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Note: question solve by using the quadratic formula

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Answer 2

Answer:

Answer is 11 and 13

Explanation:

Let us consider one odd number as 'x', so the consecutive odd number will be (x+2)

Equation : $x^2+(x+2)^2=290$

=> $x^2+x^2+4+4x=290$

=> $2x^2+4x-290+4=0$

=> $x^2+2x-143=0$

=> $x^2 +13x-11x-143=0$

=> $x(x+13)-11(x+13)=0$

=> $(x-11)(x+13)=0$

Therefore x = 11 & 13

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