Got stuck with this Word Problem

[onesun0000]

Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.

Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?

Here's what I did:

Let the population last year be n, so n = x² and x = √n
Last month: n + 100 = x² + 1
Next Month: n + 200 = x² ...

and i Got stuck there. I don't know where I am going ... Your help is appreciated

 

[Romsek]

Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.

Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?

Here's what I did:

Let the population last year be n, so n = x² and x = √n
Last month: n + 100 = x² + 1
Next Month: n + 200 = x² ...

and i Got stuck there. I don't know where I am going ... Your help is appreciated

$\displaystyle p_0 = k^2 + 100 = m^2 + 1$

$\displaystyle k^2 + 200 = n^2$

subtract the first equation from the second

$\displaystyle 100 = n^2 - m^2 - 1,~n>m$

$\displaystyle 101 = n^2 - m^2$

$\displaystyle 101 = (n-m)(n+m)$

you should be able to solve this w/o too much problem.

$\displaystyle \text{The factors of 101 are } 1,~101$

$\displaystyle \text{so }n-m=1,~n+m=101$

$\displaystyle n=51,~m=50$

$\displaystyle p_0 = k^2 = 51^2 - 200 = 2401$

 

[Steven G]

Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.

Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?

Here's what I did:

Let the population last year be n, so n = x² and x = √n
Last month: n + 100 = x² + 1
Next Month: n + 200 = x² ...

and i Got stuck there. I don't know where I am going ... Your help is appreciated

I'm confused. You say that n=x². Then you say that n+100 = x² + 1, ie you added 100 to the left side and 1 to the right side of n=x². Of course this is not correct.

 

[Harry_the_cat]

Here's a problem about perfect squares and it's very hard for me. I tried to solve but I got stuck.

Last year, the town of Whipple had a population that was a perfect square. Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square. Next month, 100 more people will move to Whipple, making the population a perfect square again. What was the original population of Whipple?

Here's what I did:

Let the population last year be n, so n = x² and x = √n
Last month: n + 100 = x² + 1
Next Month: n + 200 = x² ...

and i Got stuck there. I don't know where I am going ... Your help is appreciated

It helps if you know the pattern between consecutive square numbers.

1, 4, 9, 16, 25, 36, …..

Look at the differences between these consecutive squares
4 - 1 = 3
9 - 4 = 5
16 -9 = 7 etc

Your two perfect squares differ by 200 (while the roots differ by 2), so can you use the pattern to find the squares that differ by 99 and 101 (to make up the total diff of 200)?

 

[onesun0000]

Hi guys,

Thank you so much for your help. I now understand how to solve it. Hope I can help you with Math someday