Struggling w/solving rational inequalities

[lilrhino]

:?:

I'm struggling to solve this rational inequality:

x+1/x-2+x-3/x-1<0

I looked for critical values in the 1st step - 2 and 1 which shows that f(x) is not defined for x=2 and x=1.

(x+1)/(x-2) + (x-3)/(x-1) =0
I set the problem above to = 0 and then,

I got rid of the fraction by multiplying (x-2)(x-1) by both sides

Next, I got,

(x-1) (x+1) + (x-2) (x-3) = 0

After FOIL I get:
x^2 - 1 + (x^2-3x-2x+6)=0
combined terms to get this:
2x^2-5x+5=0

I get stuck on what I should do next to solve. In the problems where subtraction is involved, the x^2 gets cancelled out and I can solve for x. In this case I can't figure it out. Please help!

 

[Mrspi]

:?:

I'm struggling to solve this rational inequality:

x+1/x-2+x-3/x-1<0

I looked for critical values in the 1st step - 2 and 1 which shows that f(x) is not defined for x=2 and x=1.

(x+1)/(x-2) + (x-3)/(x-1) =0
I set the problem above to = 0 and then,

I got rid of the fraction by multiplying (x-2)(x-1) by both sides

Next, I got,

(x-1) (x+1) + (x-2) (x-3) = 0

After FOIL I get:
x^2 - 1 + (x^2-3x-2x+6)=0
combined terms to get this:
2x^2-5x+5=0

I get stuck on what I should do next to solve. In the problems where subtraction is involved, the x^2 gets cancelled out and I can solve for x. In this case I can't figure it out. Please help!

Well...you could always use the quadratic formula to solve for x...in this case, that will give you two complex numbers.

So, check the regions LESS than 1, BETWEEN 1 and 2, and GREATER than 2 to find where the original inequality is true.

 

[lilrhino]

Well...you could always use the quadratic formula to solve for x...in this case, that will give you two complex numbers.

So, check the regions LESS than 1, BETWEEN 1 and 2, and GREATER than 2 to find where the original inequality is true.

Thanks for the response!

I tried the quadratic formula and got this:
2x^2-5x+5
a = 2, b = -5, c = 5

x=-(-5)± Sq Root (-5)^2 - 5(2)(5)÷ 2(2)

5 ± Sq Root ( 25 - 40 ) ÷ 4
= 5± Sq Root (-20/4 ) = -5 or 5 ± Sq Root ( 30/4 ) or (15/2)
Am I on the right track?

 

[tkhunny]

2x^2-5x+5

Where did the equation go?

x=-(-5)± Sq Root (-5)^2 - 5(2)(5)÷ 2(2)

Serious notation problems and an error.

x={-(-5)± Sq Root [(-5)^2 - \(\displaystyle 4\)(2)(5)]}÷ 2(2)

Also, I don't see where you have remembered the original INequality.

 

[lilrhino]

lilrhino wrote:
2x^2-5x+5
Where did the equation go?
Quote:
x=-(-5)± Sq Root (-5)^2 - 5(2)(5)÷ 2(2)
Serious notation problems and an error.

x={-(-5)± Sq Root [(-5)^2 - (2)(5)]}÷ 2(2)

Also, I don't see where you have remembered the original INequality.

The original rational inequality:

(x+1)/(x-2) + (x-3)/(x-1) < 0

(x+1)/(x-2) + (x-3)/(x-1) = 0

The denominator is 0 when x=2 or x=1


As I mentioned in my original post. I'm not understanding how this works. The examples in the book show only examples where subtraction is involved so the x^2 cancels out. In this case, I'm getting 2x^2-5x+5=0 and can't move forward.

According to the text:

To solve a rational inequality:

1) find an equivalent inequality with 0 on one side (which I've done)
2) change the inequality symbol to an equals sign and solve the related equation (haven't been able to solve)
3) I found values of the variable for which the related rational function is not defined.
4) I could find the intervals after I've solved but I can't move on to Step 4.

 

[tkhunny]

Really, you're very close to already done.

2x^2 - 5x + 5 = 0

(-5)^{2} - 4*2*5 = 25 - 40 = -15

This has no Real solutions. It is ALWAYS greater than zero (0).

It may help to rewrite the expression as 2 + 3/(x-2) - 2/(x-1).

Having no REAL solution (i.e. the graph never crosses the x-axis), you need to worry only about the asymptotes that you have identified already. x = 1 and x = 2. There is a way to tell where this thing goes in the three regions so defined, but it's easy enough just to try a value. x = 0 tends to be rather easy.

For x = 0, (0+1)/(0-2) + (0-3)/(0-1) = 1 > 0 -- This does not meet the requirements of the original inequality.

For x = 10, (10+1)/(10-2) + (10-3)/(10-1) = {something} > 0 -- This does not meet the requirements of the original inequality.

You do it for 1 < x < 2 and see if you get something less than zero (0).

 

[tkhunny]

After FOIL I get:

Just for the record, the concept is MULTIPLICATION. I have little doubt that whoever invented "FOIL" regrets the vast numbers of students who have thus forgotten that they were multiplying.

 

[lilrhino]

Thanks tkhunny for your response. I figured that there was no real solution which explained why I was having so many problems solving.