# Thread: inequalities: 2 methods

1. ## inequalities: 2 methods

Hi guys,

I am trying to find the values of x for which

$\displaystyle x - 3 < \frac{x - 4}{x}$

Attached is a diagram from which it is easy to see the answer x < 0! i.e these are the values of x for which the straight line x - 3 is below the curve (which I have expressed as:

$\displaystyle 1 - \frac{4}{x}$

But, as we all know, inequalities can be solved in three ways: by sketching, by calculation and by completing the square. So I have tried it by calculation - here goes:

$\displaystyle x - 3 < (x - 4) / x$

$\displaystyle x^2 - 3x < x - 4$

$\displaystyle x^2 - 4x + 4 < 0$

$\displaystyle (x - 2) (x - 2) < 0$

$\displaystyle (x - 2)^2 < 0$

Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!

2. Originally Posted by s_ingram
...So I have tried it by calculation - here goes:

$\displaystyle x - 3 < (x - 4) / x$

$\displaystyle x^2 - 3x < x - 4$

$\displaystyle x^2 - 4x + 4 < 0$

$\displaystyle (x - 2) (x - 2) < 0$

$\displaystyle (x - 2)^2 < 0$

... Can anyone see what I am doing wrong!
You assumed that $\displaystyle x \geq 0$. If you do the calculations for x < 0 you'll get the result which you already know.

3. Originally Posted by s_ingram
$\displaystyle x - 3 < (x - 4) / x$

$\displaystyle x^2 - 3x < x - 4$

$\displaystyle x^2 - 4x + 4 < 0$

$\displaystyle (x - 2) (x - 2) < 0$

$\displaystyle (x - 2)^2 < 0$

Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!
I wouldn't multiply both sides by x. Instead, I would do this:
\displaystyle \begin{aligned} x - 3 &< \frac{x - 4}{x} \\ x - 3 - \frac{x - 4}{x} &< 0 \\ \frac{x^2 - 3x}{x} - \frac{x - 4}{x} &< 0 \\ \frac{x^2 - 4x + 4}{x} &< 0 \\ \frac{(x - 2)^2}{x} &< 0 \end{aligned}

At this point I would make a sign chart. First, I need "critical points." These are values which (1) the fraction equals 0, and (2) the fraction is undefined. I find (1) by setting the numerator equal to zero:
\displaystyle \begin{aligned} (x - 2)^2 &= 0 \\ x - 2 &= 0 \\ x &= 2 \end{aligned}
I find (2) by setting the denominator equal to zero: x = 0.

Then I draw my sign chart like this:
Code:
         und              0
----------+---------------+-------------
0               2
(und = undefined)

Now take the fraction
$\displaystyle \frac{(x - 2)^2}{x}$ and test a value less than 0. You will see that the fraction is negative, so I make a notation on my sign chart.
Code:
  (-)^2
-----
(-)   und              0
----------+---------------+-------------
neg    0               2
Pick a number between 0 and 2 and test the fraction, and then pick a number greater than 2 and test the fraction. When you're done, you should get this:
Code:
  (-)^2          (-)^2          (+)^2
-----          -----          -----
(-)    und     (+)      0     (+)
----------+---------------+-------------
neg    0      pos      2     pos
We want a range of numbers where the fraction is less than 0, or negative. So the solutions are x < 0.

01

4. Sorry, where do I assume that $\displaystyle x \geq 0$? If x is less than zero say -5, then $\displaystyle (x - 2)^2$ gives $\displaystyle (-7)^2 = 49$ and this is not less than 0.

5. Thanks again yeongil. I can see your method works and I have used something like it when checking the ranges of functions for which they are real or unreal, but I don't see why you would not multiply by x. Is it wrong to do so and if so why?

6. To answer both of your posts, you assumed that $\displaystyle x \geq 0$ when you multiplied both sides by x. Remember that when you multiply both sides of an inequality by a negative number, you flip the sign. When you went from this step:
$\displaystyle x - 3\;{\color{red}<}\;\frac{x - 4}{x}$
to this step:
$\displaystyle x^2 - 3x\;{\color{red}<}\;x - 4$
you multiplied both sides by x, and since you didn't change the sign, you assumed that $\displaystyle x \geq 0$.

The problem was to solve for x, and if we pretend that we do not know what x is beforehand, we shouldn't multiply both sides by x; instead, subtract the fraction from both sides and proceed like I did earlier.

01

7. Excellent answer! Now I see. I suspected Earboth was on to something but I didn't understand what he meant.