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

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