Inequality problem

Printable View

May 22nd 2009, 09:49 AM
craig

Inequality problem

Find the set of values of x for which:

$\frac{x}{x-3} > \frac{1}{x-2}$ .

I rearranged it to form:

$\frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0$

Therefore critical points of the denominator are x = 2, 3.

It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".

I am not sure then how to work out a set of values for x, they have x<2 and x>3 in the book.

Thanks in advance for the help :)
May 22nd 2009, 10:08 AM
Plato

Quote:

Originally Posted by craig

Find the set of values of x for which:
$\frac{x}{x-3} > \frac{1}{x-2}$ .
I rearranged it to form:
$\frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0$
Therefore critical points of the denominator are x = 2, 3.
It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".

The roots of the numerator are complex numbers.
The value of the numerator at x=0 is 3. So it is always positive.
May 22nd 2009, 10:14 AM
craig

Thanks for the reply Plato. Do you have any idea how they worked out their set of values?
May 22nd 2009, 10:27 AM
Moo

Hello,

Quote:

Originally Posted by craig

Find the set of values of x for which:

$\frac{x}{x-3} > \frac{1}{x-2}$ .

I rearranged it to form:

$\frac{x^2 - 3x + 3}{(x-3)(x-2)} > 0$

Therefore critical points of the denominator are x = 2, 3.

It says that the numerator is always positive, could you explain why exactly this is, or is this just something that you "know".

If the discriminant of a quadratic $ax^2+bx+c$ is negative, then it keeps a constant sign, the same as a.
So here, since a=1, and the discriminant is $9-12=-3$ , the above is always positive.

You can prove it by taking the derivative. For this particular polynomial.

Quote:

I am not sure then how to work out a set of values for x, they have x<2 and x>3 in the book.

Thanks in advance for the help :)

Then, for the quotient to be positive, you need the denominator to be positive.

The product of two terms is positive if and only if both terms have the same sign.
That is to say [x-2>0 and x-3>0] or [x-2<0 and x-3<0]
From the first one, you have [x>2 and x>3], which is [x>3]
From the second one, you have [x<2 and x<3], which is [x<2]

Does it look clear to you ?
May 22nd 2009, 03:05 PM
craig

Quote:

Originally Posted by Moo

If the discriminant of a quadratic $ax^2+bx+c$ is negative, then it keeps a constant sign, the same as a.

That makes perfect sense thank you, if the discriminant is negative - ie two complex roots - then the quadratic keeps the sign of the $x^2$ constant.

Never had it explained like this before, but seems much easier.

Quote:

Originally Posted by Moo

Then, for the quotient to be positive, you need the denominator to be positive.

The product of two terms is positive if and only if both terms have the same sign.
That is to say [x-2>0 and x-3>0] or [x-2<0 and x-3<0]
From the first one, you have [x>2 and x>3], which is [x>3]
From the second one, you have [x<2 and x<3], which is [x<2]

Does it look clear to you ?

Yep that's perfectly clear thank you.
May 22nd 2009, 03:07 PM
craig

Sorry just one more question, if the original question had something that was less than zero, would you do the opposite of the above, make one negative and one positive?
May 22nd 2009, 10:04 PM
Singaporean

Not sure if anyone mentioned this but, you can complete the square for the numerator to prove that it is always positive cause i dont understand the discriminant thingy always.

$x^2-3x+3=[x-\frac{3}{2}]^2+\frac{3}{4}$

Since $[x-\frac{3}{2}]^2\geq0$ for all values of x
Therefore, $[x-\frac{3}{2}]^2+\frac{3}{4}>0$ for all values of x