x + 1 + x − 1 = 1
Find the number of real x satisfying the equation above.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
Thankyou very much sir...
I couldn't understand why the squaring technique failed to produce the right result even though everything done was systematically right....
Log in to reply
The squaring technique, particularly when applied twice in succession, often produces extraneous answers, which is why we always have to check the potential solutions back in the original equation, but I can't see why this particular (non-viable) value popped out. If we had started by rewriting the given equation as x + 1 = 1 − x − 1 and used the squaring technique we would have ended up with
x + 1 = 1 + ( x − 1 ) − 2 x − 1 ⟹ 1 = − 2 x − 1 ⟹ − 2 1 = x − 1 ,
which clearly has no solutions as x − 1 ≥ 0 , but it does give some insight into how the potential solution x = 5 / 4 emerges. When the squaring technique doesn't work it does so in mysterious ways. :)
If n − 1 − n + 1 is rational, n − 1 − n + 1 = b a ; ( a , b ) = 1 , b = 0 ; a , b ∈ Q ... 1 ⟹ n − 1 − n + 1 1 = a b ⟹ a b = ( n − 1 − n + 1 ) ( n − 1 − n + 1 ) n − 1 + n + 1 ⟹ a b = ( n + 1 ) − ( n − 1 ) n − 1 + n + 1 ⟹ a b = 2 n − 1 + n + 1 ⟹ n − 1 + n + 1 = a 2 b ... 2 1 + 2 ⟹ b a + a 2 b = n − 1 − n + 1 + n − 1 − n + 1 ⟹ 2 a b a 2 + 2 b 2 = 2 n + 1 ... 3 1 − 2 ⟹ b a − a 2 b = n − 1 − n + 1 − ( n − 1 − n + 1 ) ⟹ 2 a b a 2 − 2 b 2 = n − 1 ... 4 From 3 and 4 n − 1 and n + 1 are rational, as ’a’ and ’b’ are rational. ⟹ n − 1 and n + 1 are perfect squares but, there is no possibility for a square no to differ by 2. Therefore, there is no real solutions for that equation making it rational.
http://brilliant.com/problems/rationality-2
I suppose the signs under the radical are wrong for derivation of equation 3 & 4. By the way nice line of reasoning..
x + 1 + x − 1 x + 1 − x − 1 2 x + 1 − x − 1 = 1 x + 1 = 2 3 = 2 = 2
But this causes x + 1 > 1 , so there are no solutions.
How did you get x = √3/2.. I thought it would be 5/4...
Problem Loading...
Note Loading...
Set Loading...
First note that for x − 1 to be real we require that x ≥ 1 . But this implies that x + 1 ≥ 2 > 1 , and thus there are 0 real numbers x that satisfy the equation.
Note that if we were to have tried squaring both sides we would have found that
x + 1 + ( x − 1 ) + 2 ( x + 1 ) ( x − 1 ) = 1 ⟹ 4 ( x 2 − 1 ) = ( 1 − 2 x ) 2 ⟹ − 4 = 1 − 4 x ⟹ x = 4 5 .
But with this value of x we find that x + 1 + x − 1 = 2 , i.e., in this case the "square both sides" approach yields just one potential solution which we must discard.