An algebra problem by Puneet Pinku

Algebra Level 3

x + 1 + x 1 = 1 \large \sqrt{x+1} + \sqrt{x-1} = 1

Find the number of real x x satisfying the equation above.


The answer is 0.

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3 solutions

First note that for x 1 \sqrt{x - 1} to be real we require that x 1 x \ge 1 . But this implies that x + 1 2 > 1 \sqrt{x + 1} \ge \sqrt{2} \gt 1 , and thus there are 0 \boxed{0} real numbers x 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 = 5 4 x + 1 + (x - 1) + 2\sqrt{(x + 1)(x - 1)} = 1 \Longrightarrow 4(x^{2} - 1) = (1 - 2x)^{2} \Longrightarrow -4 = 1 - 4x \Longrightarrow x = \dfrac{5}{4} .

But with this value of x x we find that x + 1 + x 1 = 2 \sqrt{x + 1} + \sqrt{x - 1} = 2 , i.e., in this case the "square both sides" approach yields just one potential solution which we must discard.

Thankyou very much sir...

Puneet Pinku - 4 years, 9 months ago

I couldn't understand why the squaring technique failed to produce the right result even though everything done was systematically right....

Puneet Pinku - 4 years, 9 months ago

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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 \sqrt{x + 1} = 1 - \sqrt{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 1 2 = x 1 x + 1 = 1 + (x - 1) - 2\sqrt{x - 1} \Longrightarrow 1 = -2\sqrt{x - 1} \Longrightarrow -\dfrac{1}{2} = \sqrt{x - 1} ,

which clearly has no solutions as x 1 0 \sqrt{x - 1} \ge 0 , but it does give some insight into how the potential solution x = 5 / 4 x = 5/4 emerges. When the squaring technique doesn't work it does so in mysterious ways. :)

Brian Charlesworth - 4 years, 9 months ago
Viki Zeta
Aug 30, 2016

If n 1 n + 1 is rational, n 1 n + 1 = a b ; ( a , b ) = 1 , b 0 ; a , b Q ... 1 1 n 1 n + 1 = b a b a = n 1 + n + 1 ( n 1 n + 1 ) ( n 1 n + 1 ) b a = n 1 + n + 1 ( n + 1 ) ( n 1 ) b a = n 1 + n + 1 2 n 1 + n + 1 = 2 b a ... 2 1 + 2 a b + 2 b a = n 1 n + 1 + n 1 n + 1 a 2 + 2 b 2 2 a b = 2 n + 1 ... 3 1 2 a b 2 b a = n 1 n + 1 ( n 1 n + 1 ) a 2 2 b 2 2 a b = 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. \text{If } \sqrt[]{n-1} - \sqrt[]{n+1} \text{ is rational, } \\ \sqrt[]{n-1} - \sqrt[]{n+1} = \frac{a}{b} ; (a, b) = 1, b \neq 0; a, b \in Q \text{ ... } \fbox{1}\\ \implies \dfrac{1}{\sqrt[]{n-1} - \sqrt[]{n+1}} = \frac{b}{a} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{(\sqrt[]{n-1} - \sqrt[]{n+1})(\sqrt[]{n-1} - \sqrt[]{n+1})} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{(n+1) - (n-1)} \\ \implies \frac{b}{a} = \dfrac{\sqrt[]{n-1} + \sqrt[]{n+1}}{2} \\ \implies \sqrt[]{n-1} + \sqrt[]{n+1} = \dfrac{2b}{a} \text{ ... } \fbox{2} \\ \\ \fbox{1} + \fbox{2} \\ \implies \frac{a}{b} + \frac{2b}{a} = \sqrt[]{n-1} - \sqrt[]{n+1} + \sqrt[]{n-1} - \sqrt[]{n+1} \\ \implies \dfrac{a^2+2b^2}{2ab} = 2\sqrt[]{n+1} \text{ ... } \fbox{3} \\ \fbox{1} - \fbox{2} \\ \implies \frac{a}{b} - \frac{2b}{a} = \sqrt[]{n-1} - \sqrt[]{n+1} - (\sqrt[]{n-1} - \sqrt[]{n+1}) \\ \implies \dfrac{a^2-2b^2}{2ab} = \sqrt[]{n-1} \text{ ... } \fbox{4} \\ \\ \text{From } \fbox{3} \text{ and } \fbox{4} \\ \sqrt[]{n-1} \text{ and } \sqrt[]{n+1} \text{ are rational, as 'a' and 'b' are rational.}\\ \implies n-1 \text{ and } n+1 \text{ are perfect squares} \\ \text{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..

Puneet Pinku - 4 years, 9 months ago
Sharky Kesa
Aug 30, 2016

x + 1 + x 1 = 2 2 x + 1 x 1 = 2 x + 1 x 1 = 1 x + 1 = 3 2 \begin{aligned} \sqrt{x+1}+\sqrt{x-1}&=2\\ \dfrac {2}{\sqrt{x+1}-\sqrt{x-1}}&=2\\ \sqrt{x+1}-\sqrt{x-1}=1\\ \sqrt{x+1}=\frac{3}{2}\\ \end{aligned}

But this causes x + 1 > 1 \sqrt{x+1}>1 , so there are no solutions.

How did you get x = √3/2.. I thought it would be 5/4...

Puneet Pinku - 4 years, 9 months ago

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Sorry! Typo!

Sharky Kesa - 4 years, 9 months ago

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