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# Help me!

$(x+1)( x^{2}+1)=( \sqrt{x+1}+1)(x+2)$

Note by Cong Thanh
7 months, 1 week ago

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What have you tried? Have you tried squaring the equation? The only real solution is $$\dfrac{1+\sqrt5}2$$. · 7 months, 1 week ago

Can you solve it in steps? It also has $$(1- \sqrt{5})/2$$ · 7 months, 1 week ago

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: $$\dfrac{1\pm\sqrt5}2$$. · 7 months, 1 week 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 $$f(a)=(a^{2}+1)(a+1)$$ <=> $$x= \sqrt{x+1} => x^{2}-x-1=0$$ · 7 months, 1 week 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? · 7 months, 1 week ago

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 $$-ve$$.)

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 ? · 7 months, 1 week ago

You don't need to use quadratic discriminant. Other than that, your solution is perfect. · 7 months, 1 week ago

Are you talking abt alternate method to prove one-one nature of $$f(x)$$? · 7 months, 1 week ago

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.... · 7 months, 1 week ago

Oh that way... Oops :-) · 7 months, 1 week ago