y=x+1 "works" if you take the negative sign in the square roots on the right hand side (or in the terms in y). Squaring-up "forgets" the sign of the roots, that's why it gives y=x+1 as a formal solution.
Let x=(sinh α)^2 and y=(cosh β)^2(α, β≧0), the given equation can be written as follows: sinh α-cosh β=cosh α-sinh β ⇔e^α+e^β=0 There is NO solutions(∵e^α, e^β>0).
Not clear what he was trying to achieve since no real solution was fairly obvious. Maybe demonstrating that squaring both sides introduces relations that may not hold in the original and must be checked.
👍 x >/= 0 , y >/= 1 y = x + 1 x < x + 1 => sqrt( x ) - sqrt( x + 1 ) < 0 => sqrt( x + 1 ) - sqrt( x ) > 0 sqrt( x ) - sqrt( x + 1 ) =/= sqrt( x + 1 ) - sqrt( x ) x >/= 0 No soluzioni !
y=x+1 "works" if you take the negative sign in the square roots on the right hand side (or in the terms in y). Squaring-up "forgets" the sign of the roots, that's why it gives y=x+1 as a formal solution.
I love it when you correct with studs
Thanks for the video.
Let x=(sinh α)^2 and y=(cosh β)^2(α, β≧0),
the given equation can be written as follows:
sinh α-cosh β=cosh α-sinh β
⇔e^α+e^β=0
There is NO solutions(∵e^α, e^β>0).
Not clear what he was trying to achieve since no real solution was fairly obvious. Maybe demonstrating that squaring both sides introduces relations that may not hold in the original and must be checked.
👍
x >/= 0 , y >/= 1
y = x + 1
x < x + 1 => sqrt( x ) - sqrt( x + 1 ) < 0
=> sqrt( x + 1 ) - sqrt( x ) > 0
sqrt( x ) - sqrt( x + 1 )
=/=
sqrt( x + 1 ) - sqrt( x )
x >/= 0
No soluzioni !
y = x+1