# Problem with inequalities

• Jan 30th 2009, 06:02 PM
nathan02079
Problem with inequalities
Well, I'm not done yet. Am I on the right track?

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

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

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

$\displaystyle (x-7)(x-3)\leq{0}$
$\displaystyle 0\leq{(x-4)(x+3)}$
• Jan 30th 2009, 06:33 PM
mollymcf2009
Hi Nathan02079!

Are you supposed to be solving the inequality for x? Or just evaluating the function until you find what is $\displaystyle \leq 0 \leq$ ?
• Jan 30th 2009, 06:34 PM
nathan02079
I am solving for x.
• Jan 30th 2009, 07:00 PM
mollymcf2009
Looks good to me!
• Feb 1st 2009, 05:58 AM
pankaj
Quote:

Originally Posted by nathan02079
Well, I'm not done yet. Am I on the right track?

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

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

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

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

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.

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

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

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

On taking the intersection of the above conditions you will get

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

Now the solutions you obtain by solving

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

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

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

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

Therefore the answer to the given question must be [5,7]
• Feb 1st 2009, 08:06 AM
vg284
i think answer is [4 7].
• Feb 1st 2009, 08:32 AM
pankaj
Take $\displaystyle x=4.5$ and put it in $\displaystyle x^2-9x+20$,you will obtain $\displaystyle -0.25$which is impossible since $\displaystyle \sqrt{-0.25}$ is not possible in real numbers
• Feb 8th 2009, 11:26 AM
ekledes
Therefore the answer to the given question must be {4}u[5,7]
• Feb 8th 2009, 03:43 PM
pankaj
That's correct.