Question

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

Answer 1

Answer :

13 and 14.

Explaination :-

$\begin{gathered} \sf x {}^{2}  + (x + 1) {}^{2}  = 365 \\ \sf \implies \: x {}^{2}  + x {}^{2}  + 1 + 2x = 365 \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies \: x {}^{2}  + x – 182 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf \implies x {}^{2} + 14x – 13x – 182 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf \implies x(x + 14) -13(x + 14) = 0 \\ \sf \implies(x + 14)(x – 13) = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \end{gathered}$

Thus, either, x + 14 = 0 or x – 13 = 0

$\sf \implies x = – 14 \: or x = 13$

since, the integers are positive, so x can be 13, only.

$\sf \: ∴ x + 1 = 13 + 1 = 14$

$\boxed{\sf {\pink{Thus ,two \: consecutive \: positive \: integers \: will \: be \: 13 \: and \: 14. }}}$

Answer 2

Answer:

13 and 14

Step-by-step explanation:

Let the two consecutive positive integers be x and x+1

Then,

x² +(x+1)² =365

⇒x² + x² +2x+1=365

⇒2x² +2x−364=0

⇒x² +x−182=0

Using the quadratic formula, we get

x=

$- 1 + \sqrt{1 + 728} by \: 2$

= -1 + 27

2

⇒x=13 and x=−14

But x is given to be a positive integer. ∴x not equal to −14.

Hence, the two consecutive positive integers are 13 and 14.

$hope \: \: it \: \: helps \: \: you$

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