How To solve This Perfect Square Word Problem
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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 = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
New contributor
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add a comment |
$begingroup$
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 = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
New contributor
$endgroup$
add a comment |
$begingroup$
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 = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
New contributor
$endgroup$
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 = x^2 and x = √n
Last month: n + 100 = x^2 + 1
Next Month: n + 200 = x^2 ...
and i Got stuck there. I don't know where I am going ... Your help is appreciated
algebra-precalculus square-numbers
algebra-precalculus square-numbers
New contributor
New contributor
New contributor
asked 3 hours ago
harpey1111harpey1111
111
111
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3 Answers
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votes
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
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how did you get "divided by 2"?
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– harpey1111
2 hours ago
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@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
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– Will Jagy
2 hours ago
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$
$endgroup$
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
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active
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votes
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
add a comment |
$begingroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
$endgroup$
Be careful with your variable names; it will clarify things for you. You've used $x$ to mean three different things!
Let $n=x^2$; then $n+200=y^2$, so that $x^2+200 = y^2$. Rewriting gives $y^2-x^2=(y-x)(y+x)=200$. Can you make progress from there?
answered 3 hours ago
rogerlrogerl
17.6k22746
17.6k22746
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
add a comment |
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
$begingroup$
I kind of get it now but I still don't know how to use it to solve the problem because the "Last month, 100 enlightened people moved to Whipple, making the population one more than a perfect square." makes me really confused. I'm so sorry.
$endgroup$
– harpey1111
2 hours ago
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
$endgroup$
$$ x^2 + 99 = y^2 $$
$$ x^2 + 200 = z^2 $$
$$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
The choices for $x$ are
$$ frac{99-1}{2} , ; ; ; frac{33-3}{2} , ; ; ; frac{11-9}{2} , ; ; ; $$
Then confirm that $x^2 + 99$ is really a square, and check whether $x^2 + 200$ is a square, for the three possible $x$ values.
edited 2 hours ago
answered 2 hours ago
Will JagyWill Jagy
103k5101200
103k5101200
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
how did you get "divided by 2"?
$endgroup$
– harpey1111
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
$begingroup$
@harpey1111 what are $y$ and $x$ when $y+x=99$ and $y-x=1?$
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$
$endgroup$
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$
$endgroup$
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$
$endgroup$
I like where Will Jagy starts.
$$x^2+99 = y^2\
x^2 + 200 = z^2$$
to continue I would subtract one from the other
$$z^2 - y^2 = 101\(z+y)(z-y) = 101$$
$101$ is prime
$$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$
answered 2 hours ago
Doug MDoug M
44.6k31854
44.6k31854
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
add a comment |
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
$begingroup$
good...........
$endgroup$
– Will Jagy
2 hours ago
add a comment |
harpey1111 is a new contributor. Be nice, and check out our Code of Conduct.
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