inequalities: 2 methods

**s_ingram** · July 9th 2009, 12:59 AM

Hi guys,

I am trying to find the values of x for which

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

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:

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

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:

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

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

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

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

$(x - 2)^2 < 0$

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!

**earboth** · July 9th 2009, 01:17 AM

Originally Posted by s_ingram

...So I have tried it by calculation - here goes:

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

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

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

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

$(x - 2)^2 < 0$

... Can anyone see what I am doing wrong!

You assumed that $x \geq 0$ . If you do the calculations for x < 0 you'll get the result which you already know.

**yeongil** · July 9th 2009, 01:22 AM

Originally Posted by s_ingram

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

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

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

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

$(x - 2)^2 < 0$

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:
$\begin{aligned} x - 3 &< \frac{x - 4}{x} \\ x - 3 - \frac{x - 4}{x} &< 0 \\ \frac{x^2 - 3x}{x} - \frac{x - 4}{x} &< 0 \\ \frac{x^2 - 4x + 4}{x} &< 0 \\ \frac{(x - 2)^2}{x} &< 0 \end{aligned}$

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:
$\begin{aligned} (x - 2)^2 &= 0 \\ x - 2 &= 0 \\ x &= 2 \end{aligned}$
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
$\frac{(x - 2)^2}{x}$ 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

**s_ingram** · July 9th 2009, 01:30 AM

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

**s_ingram** · July 9th 2009, 01:43 AM

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?

**yeongil** · July 9th 2009, 01:58 AM

To answer both of your posts, you assumed that $x \geq 0$ 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:
$x - 3\;{\color{red}<}\;\frac{x - 4}{x}$
to this step:
$x^2 - 3x\;{\color{red}<}\;x - 4$
you multiplied both sides by x, and since you didn't change the sign, you assumed that $x \geq 0$ .

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

**s_ingram** · July 9th 2009, 02:11 AM

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

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