How To solve This Perfect Square Word Problem












2












$begingroup$


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







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harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    2












    $begingroup$


    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







    share|cite|improve this question







    New contributor




    harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$















      2












      2








      2





      $begingroup$


      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







      share|cite|improve this question







      New contributor




      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      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






      share|cite|improve this question







      New contributor




      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      asked 3 hours ago









      harpey1111harpey1111

      111




      111




      New contributor




      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.





      New contributor





      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      harpey1111 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

          votes


















          4












          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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



















          2












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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



















          2












          $begingroup$

          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



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            2 hours ago











          Your Answer





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          3 Answers
          3






          active

          oldest

          votes








          3 Answers
          3






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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
















          4












          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            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














          4












          4








          4





          $begingroup$

          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?






          share|cite|improve this answer









          $endgroup$



          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?







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          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


















          • $begingroup$
            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











          2












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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
















          2












          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          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.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            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














          2












          2








          2





          $begingroup$

          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          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.






          share|cite|improve this answer











          $endgroup$



          $$ x^2 + 99 = y^2 $$
          $$ x^2 + 200 = z^2 $$
          $$ (y+x)(y-x) = 99 = 99 cdot 1 = 33 cdot 3 = 11 cdot 9$$
          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          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


















          • $begingroup$
            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











          2












          $begingroup$

          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



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            2 hours ago
















          2












          $begingroup$

          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



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            2 hours ago














          2












          2








          2





          $begingroup$

          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



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$






          share|cite|improve this answer









          $endgroup$



          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



          $$z+y = 101\z-y = 1\z = 51\y = 50\x = sqrt{50^2 - 200} = 49$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 hours ago









          Doug MDoug M

          44.6k31854




          44.6k31854












          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            2 hours ago


















          • $begingroup$
            good...........
            $endgroup$
            – Will Jagy
            2 hours ago
















          $begingroup$
          good...........
          $endgroup$
          – Will Jagy
          2 hours ago




          $begingroup$
          good...........
          $endgroup$
          – Will Jagy
          2 hours ago










          harpey1111 is a new contributor. Be nice, and check out our Code of Conduct.










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          harpey1111 is a new contributor. Be nice, and check out our Code of Conduct.













          harpey1111 is a new contributor. Be nice, and check out our Code of Conduct.












          harpey1111 is a new contributor. Be nice, and check out our Code of Conduct.
















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