Solution to $\sqrt{x^2-5}+3>|x-1|$

Question

I tried many ways to solve this but I just can't figure it out...
$$\sqrt{x^2-5}+3>|x-1|$$

Answer 1

Square both sides, keeping always into account that we must have $x^2\ge5$: $$ x^2-5+6\sqrt{x^2-5}+9>x^2-2x+1 $$ which becomes $$ 6\sqrt{x^2-5}>-2x-3 $$ This splits into two parts: if $-2x-3<0$, the inequality is satisfied; otherwise we can square again $$ 36(x^2-5)>4x^2+12x+9 $$ that becomes $$ 32x^2-12x-189>0 $$ The roots of the polynomial are $$ \frac{6+78}{32}=\frac{21}{8},\qquad \frac{6-78}{32}=-\frac{9}{4} $$ So we get the two cases: $$ (1)\quad \begin{cases} x\le-\sqrt{5}\text{ or }x\ge\sqrt{5}\\ x>-3/2 \end{cases} \qquad\qquad (2)\quad \begin{cases} x\le-\sqrt{5}\text{ or }x\ge\sqrt{5}\\ x\le-3/2\\ x<-9/4\text{ or }x>21/8 \end{cases} $$ and it's easy to deduce that the solutions are $$ x<-\frac{9}{4}\text{ or }x\ge\sqrt{5} $$

Answer 2

HINT: Whenever you see something like $|f(x)|$ you can re-write it as $\sqrt{(f(x))^2}$

So:$$|x-1|=\sqrt{(x-1)^2}$$

Answer 3

first the domain of the inequality is $(-\infty, -\sqrt 5] \cup [\sqrt 5, \infty).$

so we will deal with the absolute value sign by breaking the function into two pieces: on $x \le -\sqrt 5$ we need to solve $\sqrt{x^2 - 5} + 3 > 1 - x$ and on $x ge \sqrt 5$ solve $\sqrt{x^2 - 5} + 3 > x - 1$

first for $x \le -\sqrt 5:$

$ \sqrt{x^2 - 5} > -x -2 $ implies $x^2 - 5 > x^2 + 4x + 4$ which says that $x < -9/4.$

now for $x \ge \sqrt 5:$

$ \sqrt{x^2 - 5} > x -4 $ implies $x^2 - 5 > x^2 -8x + 16$ which says that $x > 21/8$ but for $x = 21/8$ the right hand side $x - 4 < 0.$ that is $x = 21/8$ solves $\sqrt{x^2 - 5} > 4 - x$ instead.

combining the two the solution is $$-\infty < x < -\dfrac{9}{4} \text{ or } \sqrt 5 < x < \infty$$

p.s. every day you learn a new thing.

Answer 4

Both sides are continuous functions of $x$, so their inequality relation cannot change direction without them being equal first, so why not simply solve $$ \sqrt{x^2-5}+3=|x-1| $$ Note that $\sqrt{x^2-5}$ is only defined for $x\in\mathbb R\setminus(-\sqrt5,\sqrt5)$. The above equation has solution $x=-\tfrac{9}{4}$ for the case $|x-1|=1-x$. It has no solutions for $|x+1|=x+1$ since $\sqrt{x^2-5}=x-4$ implies $x\geq 4$ since LHS is always positive, but squaring provides the false solution $x=\frac{21}8<4$. Then we have the candidate intervals $$ (-\infty,-\tfrac{9}{4})\qquad(-\tfrac94,-\sqrt 5]\qquad[\sqrt 5,\infty) $$ Then just plug in a single number from each of those intervals to determine the direction of the inequality on each. This will reveal the correct intervals.

Answer 5

You can eliminate the square roots or absolute values provided you know the signs of the expressions under the square roots or between the absolute value signs.

What you can use : if $a,b\ge 0$, $a>b\iff a^2 >b^2$ ; if $a,b\le 0$, $a>b\iff a^2 <b^2$ ; if $a> 0 >b$, well… $a>b$.

Here it's fairly simple: first the square root is defined if and only if $x\ge \sqrt 5$ or $x\le-\sqrt 5$.

In the first case, $x >1$ so the inequation amounts to $\sqrt{x^2-5} > x- 4$. Two subcases:

• either $x<4$, and the LHS ($\ge 0$) is greater than the RHS ($<0$).
• or $x\ge 4$, and $\sqrt{x^2-5} > x- 4\iff x^2-5 > (x-4) ^2 \iff 8x >21$,, which is indeed true if $x\ge4$.

In the second case, the inequation amounts to $\sqrt{x^2-5} > -(x+2)$. Since $x<-\sqrt 5$, we know $-(x+2)>0$, so that $\sqrt{x^2-5} > -(x+2)\iff 4x <-9 $, so that $ x < -\dfrac94\bigl(<-\sqrt 5\bigr)$.

As a conclusion we have the following solutions: $$ x< -\dfrac94\quad \text{or}\quad x\ge \sqrt 5.$$