Two equations to solve simultaneously

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Homework Help Overview

The discussion revolves around solving two simultaneous linear equations involving variables c and d, expressed in terms of constants derived from square roots. The equations are set up as 2 = c(1+sqrt(2)) + d(1-sqrt(2)) and 5 = c(1+sqrt(2))^2 + d(1-sqrt(2))^2.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore methods for isolating variables and substituting expressions between the two equations. There are discussions about manipulating the equations by multiplying them by constants and rearranging terms to express one variable in terms of the other.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting specific algebraic manipulations. There is a recognition of previous mistakes in the original poster's attempts, and one participant confirms achieving the correct answer through a suggested method.

Contextual Notes

There is an emphasis on understanding the equations as linear and the need for clear working steps to identify errors in reasoning. The original poster's previous attempts are noted as having led to incorrect conclusions, prompting a reevaluation of their approach.

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Homework Statement


2 = c(1+sqrt(2)) + d(1-sqrt(2))
5 = c(1+sqrt(2))2 + d(1-sqrt(2))2

The Attempt at a Solution


[/B]
Ive tried finding an expression for c, and replacing c in the other equation, but it gives me the wrong answer... what should i do?

The answer should be c=(sqrt(2)+2)/4
 
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Show your working. We can't help you if we can't see where you might have gone wrong.
 
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with A= 1+ \sqrt{2} and B= 1- \sqrt{2}. Yes, you could solve the first equation for c, c= (2- Bd)/A, and put that into the second equation: A^2[(2- Bd)/A)+ B^2d= A(2- Bd)+ B^2d= 2A- ABd+B^2d= 5 so (B^2- AB)d= 5- 2A and d= (5- 2A)/(B^2- AB)

Personally, I would have multiplied the first equation by A to get A^2c+ ABd= 2A and then subtract that from the second equation: (B^2- AB)d= 5- 2A which immediately gives d= (5- 2A)/(B^2- AB). Similarly, multiply the first equation by B to get ABc+ B^2d= 2B and subtract that from the second equation to get (A^2- AB)c= 5- 2B so that c= (5- 2B)/(A^2- AB).

Now put A= 1+\sqrt{2} and B= 1- \sqrt{2}.
 
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Thanks! I got the correct answer using d = (5- 2A)/(B^2- AB). Turns out I've been doing a mistake before i got to this part...
 
HallsofIvy said:
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with A= 1+ \sqrt{2} and B= 1- \sqrt{2}. Yes, you could solve the first equation for c, c= (2- Bd)/A, and put that into the second equation: A^2[(2- Bd)/A)+ B^2d= A(2- Bd)+ B^2d= 2A- ABd+B^2d= 5 so (B^2- AB)d= 5- 2A and d= (5- 2A)/(B^2- AB)

Personally, I would have multiplied the first equation by A to get A^2c+ ABd= 2A and then subtract that from the second equation: (B^2- AB)d= 5- 2A which immediately gives d= (5- 2A)/(B^2- AB). Similarly, multiply the first equation by B to get ABc+ B^2d= 2B and subtract that from the second equation to get (A^2- AB)c= 5- 2B so that c= (5- 2B)/(A^2- AB).

Now put A= 1+\sqrt{2} and B= 1- \sqrt{2}.

Which, of course, should simplify to...

c = (2 + Sqrt[2])/4

d = (2 - Sqrt[2])/4
 

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