Geometric/algebraic proof of a quadratic

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SUMMARY

The discussion focuses on constructing an algebraic proof for a quadratic equation using geometric principles, specifically through the analysis of similar triangles. Participants highlight the relationships between segments such as QN, NY, and QY, and identify multiple similar triangles including OPP', OQQ', and ONQ. The key challenge lies in establishing the necessary geometric relationships and proving that certain segments are equal, particularly in relation to the circle's center, O. The conclusion emphasizes the importance of using triangle similarity and segment ratios to derive the required algebraic expressions.

PREREQUISITES
  • Understanding of geometric principles, specifically triangle similarity
  • Knowledge of algebraic manipulation, including completing the square
  • Familiarity with properties of circles and their centers
  • Ability to analyze geometric diagrams for relationships between segments
NEXT STEPS
  • Study the properties of similar triangles in geometric proofs
  • Learn how to complete the square in quadratic equations
  • Explore the concept of geometric constructions and their algebraic implications
  • Research the relationships between circle geometry and triangle similarity
USEFUL FOR

Students studying geometry and algebra, particularly those working on proofs involving quadratics and geometric relationships. This discussion is beneficial for anyone looking to enhance their understanding of algebraic proofs through geometric methods.

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Homework Statement



http://img717.imageshack.us/img717/4029/screenshot20110106at123.png

i don't know how to construct an algebraic proof from this or how to attempt it.

thank you

Homework Equations





The Attempt at a Solution

 
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For an algebraic proof you can complete the square then solve for x.
 
yes.

but i mean with the information from the drawing.

there are two similar triangles, therefore QN/2 = NY/QY. but how can one proceed here?
 
or put it in another way, can't find more similar triangles. where is the next one?
 
There are many similar triangles, such as OPP' with OQQ' and ONQ with OPM and OPN with OMQ where O is the circle center. But I don't know where to go from there yet.
 
assuming that O is the center, which means ON=OQ=OM=QP=1, => NQ=PM=1.
but how can one deduce that the cutting point of P'Q' and PQ is the origin, is there some proof?
 
How are you able to conclude that ON = OQ for example? In particular, I doubt that P'Q' hits the origin. This would only be true if (by similarity) PP' = QQ', and there doesn't seem to be any reason why that must happen.
 
right. was a wrong assumption.

i guess the vital part is to make constructions with Q'Q,PP' and 2, the only known lengths and to equate the ratios of similar triangles to finally have QY=...,QX=...
 

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