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A week or so ago I got this problem and have tried to solve it with no success.
I am supposed to find z, y and z so that xy=-6+2√2, xz=7+7√2 and yz=2-4√2.
Help!
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Hi koothrapali;
I am not sure that I am following you here. It looks like you just want to solve the 3 equations below?!
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Or is this asking for the angles in this triangle?
In which case, (ignoring the two negative sides and plowing ahead anyway) by the cosine rule
No arccos available for this value, so it looks like I've wasted my time doing this ... oh drat
wrt post #4.
What the heck are you doing up?
Take the angles if you can not sleep
That's it in a nutshell. You'd think this would have sent me to sleep but no.
Hopefully koothrapali will put us out of our misery and explain what this question is after.
Bob
Last edited by Bob (2011-06-03 14:39:41)
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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Hi Bob;
What the heck are you doing up?
Take the angles if you can not sleep.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Sorry, very sorry. I am supposed to fing x, y and z so that the product of x and y is equal to -6+2√2, the product of xz is equal to 7+7√2 and the product of yz is equal to 2-4√2.
Again very sorry for not explaining.
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There are 2 solutions:
Or
But I still do not know why the triangle and what x,y,z are.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thank you bobbym! But one question, how did you get the answers? What was the procedure?
Last edited by koothrapali (2011-06-06 15:26:09)
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Solve in the first one for x
Substituting in 2.
Solve in the last eq for z
Substitute in this:
Solve for y
Now you have y. Just plug in to one of these
to get x or z.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thank you very much for your quick answer, and your patience.
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Hi koothrapali;
Your welcome.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I found a new way to solve for y: multiply xy and yz, then divide by xz. It also works, I did it both ways. Thank you!
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Hi koothrapali;
Yes, that will work in this case! But as a general idea it should be done carefully, if at all.
Here it is obvious that xz can not be zero. Before you make the move of dividing by unknowns you should prove that they do not equal zero, and you should state that.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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