×

# Help me!

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

Note by Cong Thanh
1 year, 8 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

## Comments

Sort by:

Top Newest

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

- 1 year, 8 months ago

Log in to reply

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

- 1 year, 8 months ago

Log in to reply

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: $$\dfrac{1\pm\sqrt5}2$$.

- 1 year, 8 months ago

Log in to reply

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

- 1 year, 8 months ago

Log in to reply

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?

- 1 year, 8 months ago

Log in to reply

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 ?

- 1 year, 8 months ago

Log in to reply

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

- 1 year, 8 months ago

Log in to reply

Are you talking abt alternate method to prove one-one nature of $$f(x)$$?

- 1 year, 8 months ago

Log in to reply

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....

- 1 year, 8 months ago

Log in to reply

Oh that way... Oops :-)

- 1 year, 8 months ago

Log in to reply

I don't know but i know you can prove it!

- 1 year, 8 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...