Question

find two consecutive odd positive integers sum of whose square is 290 .
(दो क्रमागत विषम धनात्मक संख्या ज्ञात कीजिए जिनके वर्गों का योग 290 है। )​

Answer 1

Answer:

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

In the question we are given with positive integers.

Hence the number is 11 and the consecutive odd number is 13.

Hence these are the required numbers.

Step-by-step explanation:

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