Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$ I have this equation that I have to solve:

Determine all real $x$ that satisfies $\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}$

Maybe it involves means? The square roots are leaning towards it... Could someone help me?
 A: the inequality is given by
$$\sqrt{(3-x)}>1/2+\sqrt{x+1}$$ and we have $$-1\le x\le 3$$
after squaring we get
$$\frac{7}{4}-2x>\sqrt{x+1}$$
Now it must be $$\frac{7}{8}>x\geq -1$$
squaring again we have
$$x^2-2x+\frac{33}{64}>0$$
Can you proceed?
solving the last inequality we obtain $$-1\le x<\frac{1}{8}(8-\sqrt{31})$$
A: In more detail they'll you'll probably need.
For $\sqrt{3 - x}$ to exist at all in the reals it must be $3 - x \ge 0$ so $x \le 3$.  For $\sqrt{x+1}$ to exist in the reals it must be $x + 1 \ge 0$ so $x \ge -1$.  So $-1 \le x \le 3$.  Those are the only values where the term $\sqrt{3 - x} - \sqrt{x+1}$ can make any sense.  
$\sqrt{3-x} - \sqrt{x+1} > 1/2$
We want to remove the square roots.  To do that we will have to square them.  
There are three things we have to think about when we square sides. 
i) $(a + b)^2 = a^2 + 2ab + b^2$ has three terms and the middle term still involves $a$ and $b$, so $(\sqrt a + \sqrt b)^2 = a + 2\sqrt a \sqrt b + b$ will still have the roots combined.  So we probably want to isolate the roots to each side of the inequality.  If $\sqrt a + \sqrt b = c$ then $\sqrt a = c - \sqrt b$ so $\sqrt a ^2 = (c - \sqrt b)^2$ so $a = c^2 - 2c\sqrt b + b$.  We still have a square root but only one.  So we have made progress.
ii) We need to be concerned about signs.  If $a > b >0$ than $a^2 > b^2$.  But if $ 0 > a > b$ then the opposite is true.  We must "flip" the inequality sign.  If $0 > a > b$ then $a^2 < b^2$.  If $a > 0 > b$ then we don't know if $a^2 > b^2$ or $a^2 = b^2$ or $a^2 < b^2$.  Any one of those may be true.
iii) We need to be aware of "extraneous solutions".  If we have $x > \sqrt{ x - 5}$ we know $x$ is positive. If  we square to get $x^2 > x - 5$ we have now added the extraneous possibility that $x$ might be negative.  But we know it isn't.  We must take that into account.
So let's do this:
$\sqrt{3 -x} - \sqrt{x+1} > 1/2$
Let's move one of the radicals to the other side:
$\sqrt{3 - x} > 1/2 + \sqrt{x+1}$
We know $1 > 0$ and $\sqrt{x+1} \ge0$ so if we square both sides they will be positive and the inequality signs will remain as they are.
$\sqrt{x- 3}^2 > (1/2 + \sqrt{x + 1})^2$
$3 - x > 1/4 + \sqrt{x + 1} + x + 1 = x + 1 1/4 + \sqrt{x+1}$.
Note: we have just added the extraneous possibility that $x > 3$ or $x < -1$.  But we already made a note that those will be impossible.
We still have a square root to isolate:
$3 - x > x + 1 1/4 + \sqrt{x+1}$ so
$7/4 - 2x > \sqrt{x+1}$; $ \sqrt{x+1} \ge 0$ so we can square and keep the "$>$" unflipped
$(7/4 - 2x)^2 > \sqrt{x + 1}^2$
$49/16 - 7x + 4x^2 >(x + 1)$
Note: By squaring we have added the extraneous possibility that $7/4 - 2x$ might be negative.  We know that isn't true.  So we must note: $7/4 - 2x \ge 0$ so $7/8 \ge x$ . We also know $-1 \le x \le 3$ so combining those we know $-1 \le x \le 7/8$.
Let's go on:
$49/16 - 7x + 4x^2 >(x + 1)$
$33/16 - 8x + 4x^2 >0$
$33/64 - 2x + x^2 >0$
Use the quadratic equation:
$(x - \frac{2 + \sqrt{4 +33/16}}{2})(x - \frac{2 + \sqrt{4 +33/16}}{2}) > 0$
$(x - 1-\frac{\sqrt{33}}{8})(x - 1 + \frac{\sqrt{31}}{8}) > 0$
We have two terms multiplying to  a positive result.  So EITHER they are both positive or the are both negative.
If they are both positive we have:
$x > 1 + \frac{\sqrt{31}}{8}$
But we know $-1 \le x \le 7/8$ so this is impossible.
So they are both negative.  So we know:
$x < 1 - \frac{\sqrt{31}}{8} < 1 + \frac{\sqrt{31}}{8}$
$1 - \frac{\sqrt{31}}{8} \approx .304$ so $-1 < 1 - \frac{\sqrt{7}}{2} < 7/8$
We know $-1 \le x \le 7/8$ and $x < 1- \frac{\sqrt{31}}{8}$. 
So combining we know:
$-1 \le x < 1 - \frac{\sqrt{31}}{8}$.
A: Hint: $\sqrt{3-x} - \sqrt{x+1} - \frac{1}{2}$ is monotonicall decreasing in the defined interval. If you can find it's root (call it $x_1$), you can say the set of solution to your inequality is $[-1,x_1)$
A: Note that


*

*$\sqrt{3-x}$ is defined only for $x\le3$

*$\sqrt{x+1}$ is defined only for $x\ge-1$

*The first square root is strictly decreasing and the second strictly increasing, so the whole LHS must be strictly decreasing (to see why, graph the functions)


Thus it remains to solve for $x$ in the case of equality, and take the interval as $[-1,x)$.
$$\sqrt{3-x}-\sqrt{x+1}=\frac12$$
$$\sqrt{3-x}=\frac12+\sqrt{x+1}$$
$$3-x=\frac14+\sqrt{x+1}+x+1$$
$$\frac74-2x=\sqrt{x+1}$$
$$\frac{49}{16}-7x+4x^2=x+1$$
$$4x^2-8x+\frac{33}{16}=0$$
$$x=1\pm\frac{\sqrt{31}}8$$
Checking, we see that the negative sign must be used for $x$, as only that produces an equality. Hence the range of $x$ is $[-1,1-\frac{\sqrt{31}}8)$.
