Solving for x in Large Right Triangle

In summary, in this conversation, the participants discuss solving for x in a large right triangle with a hypotenuse of 2000x+1125 and sides of 2000x-1125 and x. They try using the Pythagorean theorem and the quadratic formula, but struggle to get the correct answer. It is eventually determined that x can equal either 0 or 9000000, and the conversation ends with the participants understanding how to solve the problem correctly.
  • #1
BR24
23
0
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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  • #2
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?
 
  • #3
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.
 
  • #4
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?
 
  • #5
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.
 
  • #6
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?
 
  • #7
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.
 
  • #8
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?
 
  • #9
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.
 

Related to Solving for x in Large Right Triangle

What is the Pythagorean Theorem and how does it relate to solving for x in a large right triangle?

The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides. This theorem is often used to solve for missing side lengths, including x, in a large right triangle.

How do you use trigonometric ratios to solve for x in a large right triangle?

Trigonometric ratios, such as sine, cosine, and tangent, can be used to solve for x in a large right triangle. By identifying which sides and angles are known, and using the appropriate trigonometric ratio, you can set up and solve equations to find the value of x.

What are some common strategies for solving for x in a large right triangle?

Some common strategies for solving for x in a large right triangle include using the Pythagorean Theorem, trigonometric ratios, and the special right triangle ratios (30-60-90 and 45-45-90). It is also helpful to draw a diagram and label the known values before attempting to solve.

What are the units of measurement typically used when solving for x in a large right triangle?

The units of measurement used when solving for x in a large right triangle will depend on the given information and the desired answer. Common units include degrees for angles and units such as inches, feet, or meters for side lengths. It is important to pay attention to units and convert if necessary to ensure accurate calculations.

What are some real-world applications for solving for x in a large right triangle?

Solving for x in a large right triangle has many real-world applications, including in fields such as engineering, architecture, and physics. For example, engineers may need to determine the length of a support beam in a building, or architects may need to find the height of a structure. Understanding how to solve for x in a large right triangle is also useful in navigation and surveying.

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