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@Pi Han Goh
–
Thank for your help! My friend solves it by $f(x)=f( \sqrt{x+1})$$( x+1)( x^{2}+1)=( \sqrt{x+1}+1)( \sqrt{x+1}^{2}+1)$
It also is $f(a)=(a^{2}+1)(a+1)$
<=> $x= \sqrt{x+1}
=> x^{2}-x-1=0$

@Cong Thanh
–
That's a very nice approach! But the thing is, you didn't prove that they are the only solutions. It make sense that if $f(x) = f(\sqrt{x+1})$, then $x = \sqrt{x+1}$, but does that mean that they are the only solution?

@Rishabh Jain
–
No no. You're missing the point here. the domain for $x$ is $x>-1$ right? Just show that $3x^2 + 2x + 1$ is strictly positive and you're done.

Of course you can use quad discriminant as well....

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## Comments

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TopNewestWhat have you tried? Have you tried squaring the equation? The only real solution is $\dfrac{1+\sqrt5}2$.

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Can you solve it in steps? It also has $(1- \sqrt{5})/2$

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Oh right, I forgot to count the negative sign.

Rearrange the equation: $\dfrac{(x+1)(x^2+1)}{x+2} - 1 = \sqrt{x+1}$. Now square both sides of the equation and simplify it.

You will get the equation $x^6+2 x^5+x^4-3 x^3-7 x^2-8 x-3 = 0$. Factorize it and show that it only has two real roots: $\dfrac{1\pm\sqrt5}2$.

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$f(x)=f( \sqrt{x+1})$ $( x+1)( x^{2}+1)=( \sqrt{x+1}+1)( \sqrt{x+1}^{2}+1)$ It also is $f(a)=(a^{2}+1)(a+1)$ <=> $x= \sqrt{x+1} => x^{2}-x-1=0$

Thank for your help! My friend solves it byLog in to reply

$f(x) = f(\sqrt{x+1})$, then $x = \sqrt{x+1}$, but does that mean that they are the only solution?

That's a very nice approach! But the thing is, you didn't prove that they are the only solutions. It make sense that ifLog in to reply

$f(x)$.

This comes from the one-one nature of$f(x)=x^3+x^2+x+1$ Since $f(x)$ is a odd degree polynomial, for proving $f(x)$ to be one-one, it suffices to prove $f(x)$ is monotonically increasing.

$f'(x)=3x^2+2x+1>0\forall x\in\mathfrak R$ ( This comes from the fact that discriminant of $f'(x)$ is $-ve$.)

Hence $f(x_1)=f(x_2)\iff x_1=x_2\forall x_1,x_2\in\mathfrak R$.

Is it right @Pi Han Goh ?

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$f(x)$?

Are you talking abt alternate method to prove one-one nature ofLog in to reply

$x$ is $x>-1$ right? Just show that $3x^2 + 2x + 1$ is strictly positive and you're done.

No no. You're missing the point here. the domain forOf course you can use quad discriminant as well....

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