Solving for x in Large Right Triangle

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Discussion Overview

The discussion revolves around solving for the variable x in a large right triangle, where the hypotenuse and the other two sides are expressed in terms of x. Participants are attempting to apply the Pythagorean theorem to derive a quadratic equation but are encountering difficulties in their calculations and interpretations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant states the triangle has a hypotenuse of 2000x + 1125 and sides of 2000x - 1125 and x, expressing difficulty in applying the Pythagorean theorem correctly.
  • Another participant suggests starting with the equation (2000x - 1125)² + x² = (2000x + 1125)² and asks what results were obtained from the expansion.
  • A participant claims to arrive at the equation x² - 9000000x = 0 but questions its validity, particularly when using the quadratic formula.
  • Further contributions clarify the expansions of both sides of the equation, leading to a correction of earlier calculations and suggesting that x = 0 or x = 9000000 could be solutions.
  • Some participants express confusion over the calculations and seek clarification on the correct application of the quadratic formula and the implications of the results.
  • There is a recognition of errors in earlier math, with one participant admitting to forgetting to double the middle term during expansion.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct application of the quadratic formula or the validity of the solutions derived. There are multiple competing views on the calculations and interpretations of the results.

Contextual Notes

Some participants express uncertainty about the correctness of their mathematical steps and the implications of the values obtained for x in relation to the triangle's dimensions.

BR24
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A large right triangle has a hypotonuse of 2000x+1125. The other sides are 2000x-1125 and x. solve for x. I have tried this with pythagoreon theorem and solving a quadratic and what not but can't get it to work out. maybe i am just making a simple mistake but i can't get it to work out.
 
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BR24 said:
A large right triangle has a hypotonuse of 2000x+1125. The other sides are 2000x-1125 and x. solve for x. I have tried this with pythagoreon theorem and solving a quadratic and what not but can't get it to work out. maybe i am just making a simple mistake but i can't get it to work out.

Well you should get (2000x-1125)2+x2 = (2000x+1125)2 from the pythagoreon. Then expand. What did you get from your expansion?
 
i get simply x^2-9000000x=0. I am thinking this is wrong though because to use the quadratic formula would give you 9000000-9000000, which equals 0/2, or 0.
 
BR24 said:
i get simply x^2-9000000x=0. I am thinking this is wrong though because to use the quadratic formula would give you 9000000-9000000, which equals 0/2, or 0.

Well you have (2000x-1125)2+x2 = (2000x+1125)2

and (2000x-1125)2 = 4000000x2-2250000x+1265625 and (2000x+1125)2 = 4000000x2+2250000x+1265625.

How does this fall out?

Your equation above is incorrect. Further, you don't really need the quadratic equation. Anything of the form ax2 - bx = 0 = x(ax - b) so x = 0 or b/a. Also I don't think you're doing the quadratic formula correctly, you shouldn't have gotten 9000000-9000000. However 0 is a possible value for x.

Does that help?
 
wha happens to me was when i worked out the other side stuff got cancelled. ill have a go at it, take wha you said and let you know how it goes.
 
BR24 said:
wha happens to me was when i worked out the other side stuff got cancelled. ill have a go at it, take wha you said and let you know how it goes.

Well most of it should cancel out and you should get x2 - 4500000x = 0.

Is that what you got? If so what is the answer?
 
this has me very confused, even though it seems like it should be easy. when i expand from the pythagorean theorem i get x^2-9000000x=0, put that in quadratic formula and i get either 9000000 or 0. which doesn't seem right and doesn't work when i apply it to the triangle. what am i doing wrong? Can someone post up an answer just so i can tr to work through it and see where i went wrong because this is driving me nuts.
 
Queue said:
Well you have (2000x-1125)2+x2 = (2000x+1125)2

and (2000x-1125)2 = 4000000x2-2250000x+1265625 and (2000x+1125)2 = 4000000x2+2250000x+1265625.

How does this fall out?

Your equation above is incorrect. Further, you don't really need the quadratic equation. Anything of the form ax2 - bx = 0 = x(ax - b) so x = 0 or b/a. Also I don't think you're doing the quadratic formula correctly, you shouldn't have gotten 9000000-9000000. However 0 is a possible value for x.

Does that help?

You're right; I did math poorly (sorry!) and forgot to double the middle term.

You have (2000x-1125)2 = 4000000x2-4500000x+1265625 and (2000x+1125)2 = 4000000x2+4500000x+1265625.

So (2000x+1125)2 = (2000x-1125)2 + x2 gives us 4000000x2+4500000x+1265625 = 4000000x2-4500000x+1265625 + x2

Like terms cancel so we get 4500000x = -4500000x + x2 or x2 - 9000000x = 0 or x(x-9000000) = 0 or x = 0, 9000000.

Make sense?
 
es that makes sense, that is what i got. i must have done some bad math too, because i didn't think 9000000 would work when applied to the triangle,b ut i now see it does. thanks for the help.
 

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