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Find two consecutive positive integers , sum of whose squares is 365

Posted by Bhawani Gujjar 6 years, 9 months ago

CBSE > Class 10 > Mathematics

3 answers

Gaurav Seth 6 years, 9 months ago

let the no. be x and x+1
then we have x²+(x+1)²=365
2x²+2x-364=0
x²+x-182=0
x²+14x-13x-182=0
(x+14)(x-13)=0
That gives x=13 Avoiding negative value
Hence, one number is 13 and other one 14.

0Thank You

Yogita Ingle 6 years, 9 months ago

Let two consecutive positive integer be x and x + 1.
According to question
x² + (x + 1)² = 365
⇒ x² + x² + 2x + 1 = 365
⇒ 2x² + 2x - 364 = 0
⇒ x² + x - 182 = 0
Let us split middle term
x²+ 14x - 13x - 182 = 0
⇒ (x + 14) (x - 13) = 0
x = -14 or x = 13 [ - ve root being rejected ]
Hence required numbers are 13 and 14

0Thank You

Puja Sahoo 6 years, 9 months ago

13 and 14....

1Thank You

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