Help me!

(x+1)(x2+1)=(x+1+1)(x+2) (x+1)( x^{2}+1)=( \sqrt{x+1}+1)(x+2)

Note by Cong Thanh
3 years, 4 months ago

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What have you tried? Have you tried squaring the equation? The only real solution is 1+52 \dfrac{1+\sqrt5}2 .

Pi Han Goh - 3 years, 4 months ago

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Can you solve it in steps? It also has (15)/2 (1- \sqrt{5})/2

Cong Thanh - 3 years, 4 months ago

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Oh right, I forgot to count the negative sign.

Rearrange the equation: (x+1)(x2+1)x+21=x+1 \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 x6+2x5+x43x37x28x3=0x^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: 1±52 \dfrac{1\pm\sqrt5}2 .

Pi Han Goh - 3 years, 4 months ago

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@Pi Han Goh Thank for your help! My friend solves it by f(x)=f(x+1) f(x)=f( \sqrt{x+1}) (x+1)(x2+1)=(x+1+1)(x+12+1) ( x+1)( x^{2}+1)=( \sqrt{x+1}+1)( \sqrt{x+1}^{2}+1) It also is f(a)=(a2+1)(a+1) f(a)=(a^{2}+1)(a+1) <=> x=x+1=>x2x1=0 x= \sqrt{x+1} => x^{2}-x-1=0

Cong Thanh - 3 years, 4 months ago

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

Pi Han Goh - 3 years, 4 months ago

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

f(x)=x3+x2+x+1f(x)=x^3+x^2+x+1 Since f(x)f(x) is a odd degree polynomial, for proving f(x)f(x) to be one-one, it suffices to prove f(x)f(x) is monotonically increasing.

f(x)=3x2+2x+1>0xRf'(x)=3x^2+2x+1>0\forall x\in\mathfrak R ( This comes from the fact that discriminant of f(x)f'(x) is ve-ve.)

Hence f(x1)=f(x2)    x1=x2x1,x2Rf(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 Jain - 3 years, 4 months ago

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@Rishabh Jain You don't need to use quadratic discriminant. Other than that, your solution is perfect.

Pi Han Goh - 3 years, 4 months ago

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@Pi Han Goh Are you talking abt alternate method to prove one-one nature of f(x)f(x)?

Rishabh Jain - 3 years, 4 months ago

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@Rishabh Jain No no. You're missing the point here. the domain for xx is x>1x>-1 right? Just show that 3x2+2x+13x^2 + 2x + 1 is strictly positive and you're done.

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

Pi Han Goh - 3 years, 4 months ago

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@Pi Han Goh Oh that way... Oops :-)

Rishabh Jain - 3 years, 4 months ago

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@Pi Han Goh I don't know but i know you can prove it!

Cong Thanh - 3 years, 4 months ago

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