Help me!

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

#Algebra

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Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

Pi Han Goh - 5 years ago

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

Cong Thanh - 5 years 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: $\dfrac{1\pm\sqrt5}2$ .

Pi Han Goh - 5 years ago

@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 - 5 years ago

@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?

Pi Han Goh - 5 years ago

@Pi Han Goh – 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 ?

Rishabh Jain - 5 years ago

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

Pi Han Goh - 5 years ago

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

Rishabh Jain - 5 years ago

@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....

Pi Han Goh - 5 years ago

@Pi Han Goh – Oh that way... Oops :-)

Rishabh Jain - 5 years ago

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

Cong Thanh - 5 years ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$