Nonlinear Inequalities

[elizabethj]

Here is my problem

2x²-x<3

This is the work I have done so far

2x²-x-3<0

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

This is the point where I am getting a little lost.
Here is what I am assuming I am supposed to do

2x-3<0 x+1<0

x<3/2 x<-1


If that is what I am supposed to be doing though I don't understand what it would make the solution. Maybe (-infinity, 3/2)... This just doesn't seem right though. I have always had a hard time with these, and when you are supposed to use a union and when not to.

Some clarification would be much appreciated!

 

[Bob Brown MSEE]

2x²-x<3

means x-values where the graph is below y=3.
This is the same as asking when y=2xx-x-3 is negative.
note:
2xx-x-3 =2(x-1.5)(x+1)

 

[Deleted member 4993]

Here is my problem

2x²-x<3

This is the work I have done so far

2x²-x-3<0

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

This is the point where I am getting a little lost.
Here is what I am assuming I am supposed to do

2x-3<0 x+1<0

x<3/2 x<-1


If that is what I am supposed to be doing though I don't understand what it would make the solution. Maybe (-infinity, 3/2)... This just doesn't seem right though. I have always had a hard time with these, and when you are supposed to use a union and when not to.

Some clarification would be much appreciated!

Use your graphing calculator (or computer) plot y = (2x-3)*(x-1)

now what part of the curve is below x-axis (y<0)

Analytically,

y<0 when (2x-3)<0 AND (x+1)>0 → x < 3/2 AND x > -1 → -1 < x <3/2

OR

y<0 when (2x-3)>0 AND (x+1)< 0 → x > 3/2 AND x < -1 → Cannot happen

So......

 

[JeffM]

Here is my problem

2x²-x<3

This is the work I have done so far

2x²-x-3<0

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

This is the point where I am getting a little lost.
Here is what I am assuming I am supposed to do

2x-3<0 x+1<0

x<3/2 x<-1


If that is what I am supposed to be doing though I don't understand what it would make the solution. Maybe (-infinity, 3/2)... This just doesn't seem right though. I have always had a hard time with these, and when you are supposed to use a union and when not to.

Some clarification would be much appreciated!

Subhotosh Khan has shown you how to get an answer graphically. Here is a slow but perhaps intuitive analytic method.

First, solve the related equation, which you already know how to do.

\(\displaystyle Transform\ 2x^2 - x < 3\ to\ 2x^2 - x = 3.\)

\(\displaystyle 2x^2 - x = 3 \implies 2x^2 - x - 3 = 0 \implies (2x - 3)(x + 1) = 0 \implies x = \frac{3}{2}\ or\ x = -1.\) Good so far?

Next solve the given inequation.

In this case, you have a strict inequality. So, from your first step, you have now found two values that do not solve your inequality. (If the problem had been less than or equal, your first step would have found some values that do solve your problem.) Furthermore, because your function is continuous, you have also found that there are exactly two values of x that separate positive and negative ranges of the function.

So now you test. One separating value is - 1. Pick a lower value that is easy to work with, say - 2.

\(\displaystyle 2 * (-2)^2 - (- 2) = 2 * 4 + 2 = 10 \not< 3.\)

The other separating value is 3/2. Pick a higher value that is easy to work with, say 2.

\(\displaystyle 2 * (2)^2 - (2) = 2 * 4 - 2 = 6 \not< 3.\)

Now pick a value in between the two separating values, say 0.

\(\displaystyle 2 * (0)^2 - (0) = 2 * 0 = 0 < 3.\)

Your answer is \(\displaystyle \left(-1, \frac{3}{2}\right).\)

Now of course it is a lot quicker to use a graphing calculator if you have one handy. But this method may help you understand the logic.

 

[soroban]

Hello, elizabethj!

Here is my problem: .\(\displaystyle 2x^2 - x \:<\:3\)

This is the work I have done so far:

. . [\(\displaystyle 2x^2 - x - 3 \:<\:0 \quad\Rightarrow\quad (2x-3)(x+1)\:<\:0\)

This is the point where I am getting a little lost.

We have the product of two quaitities is negative.
. . This means the two factors have opposite signs.

There are two cases to consider:
. . (1) \(\displaystyle (2x-3)\) is positive and \(\displaystyle (x+1)\) is negative.
. . (2) \(\displaystyle (2x-3)\) is negative and \(\displaystyle (x+1)\) is positive.

(1) \(\displaystyle \begin{Bmatrix}2x-3 \:>\:0 & \Rightarrow & x \:>\:\frac{3}{2} \\ x+1 \:<\:0 & \Rightarrow & x \:<\:-1 \end{Bmatrix}\)
. . .This means \(\displaystyle x\) is greater than \(\displaystyle \frac{3}{2}\) and less than \(\displaystyle \text{-}1\).
. . .This is clearly impossible.

(2) \(\displaystyle \begin{Bmatrix}2x-3\:<\:0 & \Rightarrow & x \:<\:\frac{3}{2} \\ x+1 \:>\:0 & \Rightarrow & x \:>\:\text{-}1 \end{Bmatrix}\)
. . .This means \(\displaystyle x\) is less than \(\displaystyle \frac{3}{2}\) and greater than \(\displaystyle \text{-}1\).
. . .This is possible: .\(\displaystyle \text{-}1 \:<\ x\:<\:\frac{3}{2}\)

 

[HallsofIvy]

Yet another way to solve this (which also works for much more complicated inqualities): solve the equation first. That is, first solve the equation \(\displaystyle 2x^2- x= 3\). That is the same as \(\displaystyle 2x^2- x- 3= a(2x- 3)(x+ 1)= 0\). 2x- 3= 0 gives x= 3/2 and x+ 1= 0 gives x= -1.

Now, the point is this: for a continuous function to go from "-" to "+", and vice-versa, it must go through 0. So x= 3/2 and x= -1 separate "-" from "+". Those two points divide the number line into three sections:
a) x< -1. We can check one value, say x= -2. 2(-2)²- (-2)= 8+ 2= 10> 3. So 2x²- 2> 3 for all x< -1.

b) -1< x< 3/2. 0 is in that interval and 2(0)²- 0= 0< 3. So 2x²- x< 3 for all -1< x< 3/2.


c) x> 3/2. x= 2> 3/2 and 2(2)²- 2= 8- 2= 6> 3. So 2x²- x> 3 for all x> 3/2.