Q4:
Find two consecutive positive integers, sum of whose squares is 365.

Answer:

Let first number = x

Then
second number will one more so that next number wil x+1

Given
that sum of whose squares is 365.

x

^{2}+ ( x + 1)^{2}= 365
use
formula of (a +b)

^{2}= a^{2}+ 2ab +b^{2}
x

^{2}+ x^{2}+ 2x + 1 – 365 = 0
2x

^{2}+ 2x – 364 = 0
Divide
by 2 to simplify it

x

^{2}+ x – 182 = 0
factorize
it now

To understand the solution of this problem, I visited so many sites, guides, texts etc. But nowhere I could find the solution of my doubt.

ReplyDeleteMy doubt was how 2X is coming. Here you clearly solved that doubt and problem (see your explanation line no. 5. Use the formula (a+b)^2 = a^2+2ab+b^2.

Yes, you're a teacher thinks from students side..

Thanks a lot. God bless you.

Why we divided it by 2 to simplify it..can u tell this?

ReplyDeleteto take 2 common from all, which is then multiplied to 0 on other side

DeleteYes..To simplfy...so that we could factorize easily.

DeleteWe doesn't take -14 as answer but 14 comes after adding 1 to 13.

Delete-14 is neglected because question asks for positive numbers at the same time the numbers should be consecutive so 1 is added to 13 to get the next number....

The answer is coming 13 and -14.So how do you took 14 as the answer instead of -14?

ReplyDeleteThe question has asked for two consecutive POSITIVE integers.....hence from 13 & -14 we take 13 as it is positive

DeleteNo

ReplyDelete