What have you tried? Have you tried squaring the equation? The only real solution is \( \dfrac{1+\sqrt5}2 \).
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Pi Han Goh
·
9 months, 2 weeks ago

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@Pi Han Goh
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Can you solve it in steps? It also has \( (1- \sqrt{5})/2 \)
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Cong Thanh
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9 months, 2 weeks ago

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@Cong Thanh
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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: \( \dfrac{1\pm\sqrt5}2 \).
–
Pi Han Goh
·
9 months, 2 weeks ago

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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 \( f(a)=(a^{2}+1)(a+1) \)
<=> \( x= \sqrt{x+1}
=> x^{2}-x-1=0 \)
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Cong Thanh
·
9 months, 2 weeks ago

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@Cong Thanh
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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?
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Pi Han Goh
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9 months, 2 weeks ago

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@Pi Han Goh
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This comes from the one-one nature of \(f(x)\).

\[f(x)=x^3+x^2+x+1\]
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 \(-ve\).)

@Rishabh Cool
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You don't need to use quadratic discriminant. Other than that, your solution is perfect.
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Pi Han Goh
·
9 months, 2 weeks ago

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@Pi Han Goh
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Are you talking abt alternate method to prove one-one nature of \(f(x)\)?
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Rishabh Cool
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9 months, 2 weeks ago

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@Rishabh Cool
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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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Pi Han Goh
·
9 months, 2 weeks ago

## Comments

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TopNewestWhat have you tried? Have you tried squaring the equation? The only real solution is \( \dfrac{1+\sqrt5}2 \). – Pi Han Goh · 9 months, 2 weeks ago

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– Cong Thanh · 9 months, 2 weeks ago

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

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: \( \dfrac{1\pm\sqrt5}2 \). – Pi Han Goh · 9 months, 2 weeks ago

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– Cong Thanh · 9 months, 2 weeks ago

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 \( f(a)=(a^{2}+1)(a+1) \) <=> \( x= \sqrt{x+1} => x^{2}-x-1=0 \)Log in to reply

– Pi Han Goh · 9 months, 2 weeks ago

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?Log in to reply

\[f(x)=x^3+x^2+x+1\] 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 ? – Rishabh Cool · 9 months, 2 weeks ago

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– Pi Han Goh · 9 months, 2 weeks ago

You don't need to use quadratic discriminant. Other than that, your solution is perfect.Log in to reply

– Rishabh Cool · 9 months, 2 weeks ago

Are you talking abt alternate method to prove one-one nature of \(f(x)\)?Log in to reply

Of course you can use quad discriminant as well.... – Pi Han Goh · 9 months, 2 weeks ago

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– Rishabh Cool · 9 months, 2 weeks ago

Oh that way... Oops :-)Log in to reply

– Cong Thanh · 9 months, 2 weeks ago

I don't know but i know you can prove it!Log in to reply