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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 109

College Algebra

Section R.3

Suppose that m and n are positive integers with m > n.

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.

Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,717

You need to show that two of these when squared and added make the same result as the third when squared.

But which to choose? I can see that c is bigger than a. So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**e_jane_aran****Novice**- Registered: 2023-09-15
- Posts: 2

harpazo1965 wrote:

Suppose that m and n are positive integers with m > n.

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

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**e_jane_aran****Novice**- Registered: 2023-09-15
- Posts: 2

I apologize for my failure at formatting in the above. I though [ imath ] tags would work. Kindly please ignore the "\qquad" part at the beginning of each line. D'oh!

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 109

Bob wrote:

You need to show that two of these when squared and added make the same result as the third when squared.

But which to choose? I can see that c is bigger than a. So I suggest working out c^2 and write that down. Then get an expression for a^2 + b^2. If this comes to the same as c^2 then you are done.

Bob

Thank you, Bob.

*Last edited by harpazo1965 (2023-09-23 01:00:12)*

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**Bob****Administrator**- Registered: 2010-06-20
- Posts: 9,717

to both of you:

I have edited the above post so that it displays properly. The correct command is square brackets math

When a member makes a post, it is visible to anyone who logs in to the site. And anyone can respond to any post. If someone disobeys our rules then I will take action. This may involve any of the following: Issuing a warning; Editing the offending part of a post; Deleting the post; Banning for a limited period; Removing the person completely from the membership. This last means that all that person's posts get deleted too.

At the moment I can see no reason to do any of these things. Let's keep it that way.

Best wishes,

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 109

Ok. No problem. Let's get back to math.

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 109

e_jane_aran wrote:

harpazo1965 wrote:Suppose that m and n are positive integers with m > n.

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.Note: This formula can be used to find the sides of a right triangle that are integers such as 3, 4, 5; 5, 12, 13; and so on. Such triplets of integers are called Pythagorean triplets.

NOTE: Looking for the set up only.

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions. Which pair works?

If I add a^2 + c^2, the middle term cancels out. This does not leave me with much to play with.

If I do that, I am left with m^4 + n^4 twice. In other words, I get

2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.

*Last edited by harpazo1965 (2023-09-23 01:12:18)*

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**amnkb****Member**- Registered: 2023-09-19
- Posts: 23

harpazo1965 wrote:

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.NOTE: Looking for the set up only.

e_jane_aran wrote:

I suspect that they're expecting you to notice how the given side-length expressions hint of connections to perfect-square trinomiais.

If you add two of the above expressions, you get the third of the above expressions.

Which pairworks?

harpazo1965 wrote:

If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.If I do that, I am left with m^4 + n^4 twice. In other words, I get

2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.

so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)

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**harpazo1965****Member**- Registered: 2022-09-19
- Posts: 109

amnkb wrote:

If a = (m^2 - n^2), b = 2mn, and c = (m^2 + n^2), show that a, b, and c are the lengths of the sides of a right triangle.NOTE: Looking for the set up only.

e_jane_aran wrote:Which pairworks?harpazo1965 wrote:

If I add a^2 [and] c^2, the middle term cancels out. This does not leave me with much to play with.If I do that, I am left with m^4 + n^4 twice. In other words, I get

2(m^4 + n^4) = 2m^4 + 2n^4, which does not represent b^2.so maybe try a different pair of terms? (there r 2 other pairs; 1 of them works)

If I add a^2 + b^2 I get c^2. So, the addition of the two legs of this triangle

yields the value of the hypotenuse. You say?

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