Question

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

Answer 1

Answer:

The first number = 13

The other consecutive number  = 14

Step-by-step explanation:

Let the first number = x

Let the other consecutive number = x+1

According to the question :-

(x)^2 + (x+1)^2 = 365

=> x^2 + x^2 +2x +1 = 365

=> 2(x^2 + x) = 364

=> x^2 +x = 182

=> x^2 + x - 182 = 0

=> x^2 + 14x - 13x - 182 = 0

=> x(x+14) -13 (x + 14) =0

=> (x-13)(x+14) = 0

Hence, the value of x  is -14 and 13

Number can only be positive as mentioned in the question

So,

The first number = x = 13

The other consecutive number = x+1 = 13 + 1 = 14

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Answer 2

ANSWER :-

We Have To calculate the two consecutive positive integers whose squares sum is 365.

So,

Let the first positive integers be 'x'

& other be '(x+1)'

According to question -

${x}^{2} + {(x + 1)}^{2} = 365 \\ \\ = > {x}^{2} + {x}^{2} + 1 + 2x = 365 \\ \\ = > 2 {x}^{2} + 1 + 2x = 365 \\ \\ = > 2 {x}^{2} + 2x = 365 - 1 \\ \\ = > 2x(x + 1) = 364 \\ \\ = > {x}^{2} + x = \frac{364}{2} \\ \\ = > {x}^{2} + x = 182 \\ \\ = > {x}^{2} + x - 182 = 0\\ \\ = > {x}^{2} + 14x - 13x - 182 = 0 \\ \\ = > x(x + 14) - 13(x + 14) = 0 \\ \\ = > (x - 13)(x + 14) = 0 \\ \\ = > x - 13 = 0 \: \: \: \: \: and \: \: \: \: \: \: x + 14 = 0 \\ \\ = > x = 13 \: \: \: \: \: \: \: \: \: \: \: \: and \: \: \: \: \: \: x = - 14 \\ \\ = > - 14 \: is \: invalid \: because \: it \: is \: negative \: integer \: and \: we \: need \: positive \: integer \: \\ \\ \: so \: x = 13 \: is \: valid \: \\ \\ hence \: \\ first \: positive \: integer \: = 13 \: \\ and \: other \: positive \: integer \: is \: 13 + 1 \: = 14$

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