Question

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

Answer 1

Answer:

let, first number be x

then second number be x+1

now , as per the question

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

$365= x ^{2} + x ^{2} + 2x + 1$

$365 = 2x ^{2} + 2x + 1$

$2x ^{2} + 2x - 364 = 0$

$x ^{2} + x - 182 = 0$

$x ^{2} + 14x - 13x - 182 = 0$

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

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

x= -14 or x = 13

since, we need positive integer so ,

x=13 and

first number is 13 and second number is (13+1)=14

Answer 2

✧QuesTion:-

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

$\huge{\orange{\underline{\underline{\bf{SoluTion:}}}}}$

Let's

The two consecutive numbers be

'x' & 'x+1'

According To The Sum

➝ $\sf {x²+(x+1)² = 365}$

➝ $\sf {x²+x²+1²+2.x.1 = 365}$

$\boxed{\bf{(a+b)² = a²+b²+2ab}}$

➝ $\sf {2x²+1+2x = 365}$

➝ $\sf {2x²+2x+1-365 = 0}$

➝ $\sf {2x²+2x-364 = 0}$

✯ Take '2' as Common

➝ $\sf {x²+x-182 = 0}$

Now it is in the form of 'ax²+bx+c = 0'.

So, a = 1, b = 1 & c = -182

By Using $\green{\bf{Dharacharya\; Formula}}$

$\boxed{\bf{\frac{-b±\sqrt{b²+4ac}}{2a}}}$

→ $\sf {\frac{-1±\sqrt{(1)²+4×1×-182}}{2×1}}$

→ $\sf {\frac{-1±\sqrt{1+728}}{2}}$

→ $\sf {\frac{-1±\sqrt(27²)}{2}}$

→ $\sf {\frac{-1±27}{2}}$

Now The Roots are

☞ x = 13 or x = -14

As, x is a positive integer. So, x = 13

Then, x+1 = 14.

Therefore, Two Consecutive Numbers are $\fbox\red {13, 14}$

___________________

\\ CheCk: ✔️

=> 13²+14² = 365

=> 169+196 = 365

=> 365 = 365

.:. LHS = RHS

Hope It Helps You ✌️