Solving A Problem Involving Pythagorean Theorem And Polynomials

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I don't know how to solve this problem. What I've done so far is to construct 3 equations. However, I don't know how to solve those 3 equations:



Let y be the distance between the box and the ladder, and z be the length of the portion of the ladder that is beneath the top of the box:



$(x-1)^2+1=(10-z)^2$



$(y)^2+1=z^2$



$(x)^2+(1+y)^2=(10)^2$



I don't know how to proceed from here, however.






algebra-precalculus geometry euclidean-geometry factoring quartic-equations







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    up vote
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    favorite
    1












    enter image description here



    I don't know how to solve this problem. What I've done so far is to construct 3 equations. However, I don't know how to solve those 3 equations:



    Let y be the distance between the box and the ladder, and z be the length of the portion of the ladder that is beneath the top of the box:



    $(x-1)^2+1=(10-z)^2$



    $(y)^2+1=z^2$



    $(x)^2+(1+y)^2=(10)^2$



    I don't know how to proceed from here, however.






    algebra-precalculus geometry euclidean-geometry factoring quartic-equations







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite
      1









      up vote
      2
      down vote

      favorite
      1






      1





      enter image description here



      I don't know how to solve this problem. What I've done so far is to construct 3 equations. However, I don't know how to solve those 3 equations:



      Let y be the distance between the box and the ladder, and z be the length of the portion of the ladder that is beneath the top of the box:



      $(x-1)^2+1=(10-z)^2$



      $(y)^2+1=z^2$



      $(x)^2+(1+y)^2=(10)^2$



      I don't know how to proceed from here, however.






      algebra-precalculus geometry euclidean-geometry factoring quartic-equations







      share|cite|improve this question















      enter image description here



      I don't know how to solve this problem. What I've done so far is to construct 3 equations. However, I don't know how to solve those 3 equations:



      Let y be the distance between the box and the ladder, and z be the length of the portion of the ladder that is beneath the top of the box:



      $(x-1)^2+1=(10-z)^2$



      $(y)^2+1=z^2$



      $(x)^2+(1+y)^2=(10)^2$



      I don't know how to proceed from here, however.






      algebra-precalculus geometry euclidean-geometry factoring quartic-equations




      algebra-precalculus geometry euclidean-geometry factoring quartic-equations






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      share|cite|improve this question













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      edited 10 mins ago









      Michael Rozenberg

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      91.2k1584181










      asked 43 mins ago









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          By Pythagoras and similarity we obtain:
          $$fracxx-1=fracsqrt100-x^21$$ or
          $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or
          $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives:
          $$x=frac12(1+sqrt101-sqrt98-2sqrt101)$$ or
          $$x=frac12(1+sqrt101+sqrt98-2sqrt101).$$






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            By Pythagoras and similarity we obtain:
            $$fracxx-1=fracsqrt100-x^21$$ or
            $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or
            $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives:
            $$x=frac12(1+sqrt101-sqrt98-2sqrt101)$$ or
            $$x=frac12(1+sqrt101+sqrt98-2sqrt101).$$






            share|cite|improve this answer


























              up vote
              5
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              By Pythagoras and similarity we obtain:
              $$fracxx-1=fracsqrt100-x^21$$ or
              $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or
              $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives:
              $$x=frac12(1+sqrt101-sqrt98-2sqrt101)$$ or
              $$x=frac12(1+sqrt101+sqrt98-2sqrt101).$$






              share|cite|improve this answer
























                up vote
                5
                down vote










                up vote
                5
                down vote









                By Pythagoras and similarity we obtain:
                $$fracxx-1=fracsqrt100-x^21$$ or
                $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or
                $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives:
                $$x=frac12(1+sqrt101-sqrt98-2sqrt101)$$ or
                $$x=frac12(1+sqrt101+sqrt98-2sqrt101).$$






                share|cite|improve this answer














                By Pythagoras and similarity we obtain:
                $$fracxx-1=fracsqrt100-x^21$$ or
                $$x^4-2x^3-98x^2+200x-100=0,$$ where $1<x<10,$ or
                $$(x^2-x+1)^2-101(x-1)^2=0,$$ which after factoring gives:
                $$x=frac12(1+sqrt101-sqrt98-2sqrt101)$$ or
                $$x=frac12(1+sqrt101+sqrt98-2sqrt101).$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 13 mins ago

























                answered 37 mins ago









                Michael Rozenberg

                91.2k1584181




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