Question

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


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Answer 1

let one number = x

and the other one = x +1

According to question.....

x+x+1=365

2x=365-1

2x=364

x= 364 /2

x= 182

so one number =182

and the other = 182 +1 = 183

verification...

182+183

=365

Answer 2

$\color{lime}{ \underline{ \underline{ \mathfrak{ \color{teal}{Question : }}}}}$

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

$\color{lime}{ \underline{ \underline{ \mathfrak{ \color{teal}{Correct \: answer : }}}}}$

➠ 13, and 14.

$\color{lime}{ \underline{ \underline{ \mathfrak{ \color{teal}{Explanation : }}}}}$

As it is told that the numbers are consecutive then,

Assuming

➠ x

➠ (x + 1)

Are the two positive consecutive numbers.

Now,

$☯ \begin{gathered}\underline{\boldsymbol{According\: to \:the\: question :}}\\\end{gathered}$

$\rm \implies {x}^{2} +( {x + 1)}^{2} = 365$

$\rm \implies {x}^{2} + {x}^{2} + 2x + 1 = 365$

$\rm \implies {2x}^{2} + 2x + 1 - 365 = 0$

$\rm \implies {2x}^{2} + 2x - 364 = 0$

$\rm \implies {x}^{2} + x - 182= 0$

$\gray {\sf{(using \:quadratic \: fomula) }}$

$\rm \implies x = \frac{1± \sqrt{ 1 + 728} }{2}$

$\rm \implies x = \frac{ - 1 ±\sqrt{729} }{2}$

$\rm \implies x = \frac{ - 1± 27}{2}$

$\rm \implies \: x = \frac{ - 1 - 27}{2} \: or, \: \:x = \frac{ - 1 + 27}{2}$

$\rm \implies x = \frac{ \cancel{- 28}}{ \cancel 2} or, \: \: x = \frac{ \cancel{26}} { \cancel 3}$

$\bf \therefore \: x = - 14$

$or,$

$\bf \therefore x = 13$

After solving this we get :

➠ x = -14

And also,

➠ x = 13

But here x = -14 ( - ) negative integer which is not acceptable as two positive consecutive numbers.

Hence,

Required answer :

The two consecutive numbers are 13, and 14