# Thread: inequality on the set of real numbers

1. ## inequality on the set of real numbers

Solve the following inequality on the set of real numbers

$\displaystyle x^2-3\sqrt{x^2+3}\leq1$

2. Originally Posted by perash
Solve the following inequality on the set of real numbers

$\displaystyle x^2-3\sqrt{x^2+3}\leq1$
$\displaystyle x^2-3\sqrt{x^2+3}\leq1 \Rightarrow (x^2 - 1)^2 \leq 9(x^2 + 3)$
A little bit of algebra leads us to,(that is bringing everything to the LHS and factorising)
$\displaystyle (x-\sqrt{13})(x+\sqrt{13})(x^2+2) \leq 0$
$\displaystyle \forall x \in \mathbb{R}, x^2 + 2 > 0$
So,
$\displaystyle (x-\sqrt{13})(x+\sqrt{13})\leq 0$
Since exactly one of them is non-positive,
$\displaystyle x \in [-\sqrt{13},\sqrt{13}]$

3. Hello, perash!

Solve on the set of real numbers: .$\displaystyle x^2-3\sqrt{x^2+3}\:\leq\:1$

We have: .$\displaystyle x^2-1 \:\leq \:3\sqrt{x^2+3}$

Square: .$\displaystyle x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$

. . which factors: .$\displaystyle (x^2+2)(x^2-13)\:\leq\:0$

Since $\displaystyle (x^2+2)$ is always positive, $\displaystyle (x^2-13)$ must be negative.

We have: .$\displaystyle x^2-13 \:\leq \:0\quad\Rightarrow\quad x^2\:\leq\:13\quad\Rightarrow\quad |x| \:\leq\:\sqrt{13}$

Therefore: .$\displaystyle -\sqrt{13}\;\leq\: x \:\leq\:\sqrt{13}$

Drat . . . too slow again!
.

4. Originally Posted by Soroban
Hello, perash!

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We have: .$\displaystyle x^2-1 \:\leq \:3\sqrt{x^2+3}$

Square: .$\displaystyle x^4-2x^2 + 1 \:\leq \:9x^2+27\quad\Rightarrow\quad x^4 - 11x^2 - 26 \:\leq \:0$
I always get confused when I need to square an inequality.

$\displaystyle x^2 - 1 \leq 3\sqrt{x^2 + 3}$

If $\displaystyle x \in (-1, 1)$ then when we square this we shouldn't we get
$\displaystyle (x^2 - 1) \geq 9(x^2 + 3)$
since the LHS is negative? So don't we have to do the problem separately on $\displaystyle (-1, 1)$ and $\displaystyle \mathbb{R} - (-1, 1)$?

-Dan

Edit: This question has been resolved. See here.

5. Originally Posted by topsquark
since the LHS is negative? So don't we have to do the problem separately on $\displaystyle (-1, 1)$ and $\displaystyle \mathbb{R} - (-1, 1)$?

-Dan
I will try to answer this question,
First we note RHS is always positive for this problem,so if the LHS is negative then, LHS $\displaystyle \leq 0 \leq$ RHS , so it will always hold.

The point is almost all steps in the inequality reduction are equivalent.So there is no reason it will be wrong.