Two equations to solve simultaneously

  • Thread starter Thread starter johann1301
  • Start date Start date
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
johann1301
Messages
216
Reaction score
1

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
 
Last edited by a moderator:
Physics news on Phys.org
You understand that these are just linear equations, don't you: Ac+ Bd= 2 and A^2c+ B^2d= 5 with [tex]A= 1+ \sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex]. Yes, you could solve the first equation for c, [tex]c= (2- Bd)/A[/tex], 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 [tex]A= 1+\sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex].
 
  • Like
Likes   Reactions: johann1301
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 [tex]A= 1+ \sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex]. Yes, you could solve the first equation for c, [tex]c= (2- Bd)/A[/tex], 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 [tex]A= 1+\sqrt{2}[/tex] and [tex]B= 1- \sqrt{2}[/tex].

Which, of course, should simplify to...

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

d = (2 - Sqrt[2])/4