mr.t Messages 7 Reaction score 0 Thread starter Aug 18, 2008 #1 Homework Statement What is the steps between the two equations? Homework Equations [tex]P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0[/tex] Thanks!
Homework Statement What is the steps between the two equations? Homework Equations [tex]P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0[/tex] Thanks!
tiny-tim Science Advisor Homework Helper Messages 25,837 Reaction score 258 Aug 18, 2008 #2 mr.t said: What is the steps between the two equations? [tex]P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0[/tex] Thanks! Hi mr.t! i] put the a2P on the left ii] multiply both sides by (P + 1)
mr.t said: What is the steps between the two equations? [tex]P = a^{2}P+b^{2}P^{2}/(P+1) \Rightarrow P^{2}-(a^{2}+b^{2}-1)P-b^{2} = 0[/tex] Thanks! Hi mr.t! i] put the a2P on the left ii] multiply both sides by (P + 1)
mr.t Messages 7 Reaction score 0 Aug 18, 2008 #3 Hello Tiny-tim! Putting the [tex]a^{2}P[/tex] to the left and multiply both sides with (P+1) gives: [tex](P+1)(P-a^{2}P) = b^{2}P^{2} \Rightarrow[/tex] [tex]\Rightarrow P^{2}-a^{2}P^{2}+P-a^{2}P-b^{2}P^{2} = 0 \Rightarrow[/tex] [tex]\Rightarrow P^{2}(1-a^{2}-b^{2})+P(1-a^{2}) = 0[/tex] But I still can't come to the final solution? :S
Hello Tiny-tim! Putting the [tex]a^{2}P[/tex] to the left and multiply both sides with (P+1) gives: [tex](P+1)(P-a^{2}P) = b^{2}P^{2} \Rightarrow[/tex] [tex]\Rightarrow P^{2}-a^{2}P^{2}+P-a^{2}P-b^{2}P^{2} = 0 \Rightarrow[/tex] [tex]\Rightarrow P^{2}(1-a^{2}-b^{2})+P(1-a^{2}) = 0[/tex] But I still can't come to the final solution? :S
Defennder Homework Helper Messages 2,590 Reaction score 4 Aug 18, 2008 #4 I don't think the two steps are equivalent. Let b=0, then a^2=1 if [tex]P \neq 0[/tex]. Substituting these into the second equation, we have a^2 = 1 only if [tex]P = 0[/tex]
I don't think the two steps are equivalent. Let b=0, then a^2=1 if [tex]P \neq 0[/tex]. Substituting these into the second equation, we have a^2 = 1 only if [tex]P = 0[/tex]
mr.t Messages 7 Reaction score 0 Aug 18, 2008 #5 It could be an error in the solution then no fun try to learn stuff when you end up spending your time figuring out the imposible... thanks for your time anyway!
It could be an error in the solution then no fun try to learn stuff when you end up spending your time figuring out the imposible... thanks for your time anyway!