Question

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

Answer 1

Answer:

tue number are 13 and 14.

$...$

refer to the attachment for the solution.

Answer 2

• Let the first number be x
• Then, the other consecutive number will x+1

According to question,

⇒$\small\mathrm{x²+(x+1)²=365}$

⇒$\small\mathrm{x²+x²+1+2x=365}$

⇒$\small\mathrm{2x²+2x-364=0}$

Now, by splitting method :-

⇒$\small\mathrm{2x²+2x-364=0}$

⇒$\small\mathrm{2x²-28x-26x-364=0}$

⇒$\small\mathrm{2x(x-14)-26(x-14)}$

⇒$\small\mathrm{(2x-26)\:\:\:(x-14)}$

∴ $\small\mathrm{x=13\:\:\:\:and\:\:\:\:x=14}$

So, two positive consecutive integers whose sum squares will be 365 are 13 and 14.

Verification:-

⇒$\small\mathrm{13²+14²=365}$

⇒$\small\mathrm{169+196=365}$

Verified✔