Problem with inequalities

**nathan02079** · January 30th 2009, 06:02 PM

Well, I'm not done yet. Am I on the right track?

$\sqrt{x^2-9x+20}\leq\sqrt{x-1}\leq\sqrt{x^2-13}$

$x^2-9x+20\leq{x-1}$
$x-1\leq{x^2-13}$

$x^2-10x+21\leq{0}$
$0\leq{x^2-x-12}$

$(x-7)(x-3)\leq{0}$
$0\leq{(x-4)(x+3)}$

**mollymcf2009** · January 30th 2009, 06:33 PM

Hi Nathan02079!

Are you supposed to be solving the inequality for x? Or just evaluating the function until you find what is $\leq 0 \leq$ ?

**nathan02079** · January 30th 2009, 06:34 PM

I am solving for x.

**mollymcf2009** · January 30th 2009, 07:00 PM

Looks good to me!

**pankaj** · February 1st 2009, 05:58 AM

Originally Posted by nathan02079

Well, I'm not done yet. Am I on the right track?

$\sqrt{x^2-9x+20}\leq\sqrt{x-1}\leq\sqrt{x^2-13}$

$x^2-9x+20\leq{x-1}$
$x-1\leq{x^2-13}$

$x^2-10x+21\leq{0}$
$0\leq{x^2-x-12}$

$(x-7)(x-3)\leq{0}$
$0\leq{(x-4)(x+3)}$

NOT yet
First of all you need to find the values of x for which all the square roots are possible.i.e quantities under the square root signs must be non-negative.

$x^2-9x+20\geq 0,i.e.x\in(-\infty,4]\cup[5,\infty)$

$x-1\geq 0,i.e x\in[1,\infty)$

$<br /> x^2-13\geq 0,i.e x\in(-\infty,-\sqrt{13}]\cup[\sqrt{13},\infty)<br />$

On taking the intersection of the above conditions you will get

$<br /> x\in[1,4]\cup[5,\infty).<br />$
which can be said to be the domain of the inequality.

Now the solutions you obtain by solving

$(x-7)(x-3)\leq{0}$ i.e $x\in[3,7]$

$0\leq{(x-4)(x+3)}$ i.e $x\in(-\infty,-3]\cup[4,\infty)$

Therefore the values of x satisfying both the inequalities are $x\in[4,7]$

which must satisfy $x\in[1,4]\cup[5,\infty)$ .

Therefore the answer to the given question must be [5,7]