Quadratic Inequalities: Solving x² - x < 0

[MindRafter]

Solve the Inequality:
x² - x < 0

Express the solution set as intervals or union of intervals. Use the result
√a² = |a| as appropriate.

What is the procedure/explanation for the answer to this question? The answer is (0,1).
THat is: (0,1) is the solution set.
Please help.

 

[0rthodontist]

First you change the inequality into an equality:
x² - x = 0
Can you solve for x? This will divide the real numbers into intervals where the inequality uniformly holds or uniformly does not hold. Then you test each interval to find out which is which.

 

[MindRafter]

Wow, that's a fantastic solution. It didn't strike me at all...but it works. What I did before was somehting like this:
x² - x < 0
therefore x(x-1) <0
therefore x-1<0
therefore x < 1
But x-1<0,
Therefore x lies in the range 1 and 0.

Is this right??

 

[VietDao29]

Wow, that's a fantastic solution. It didn't strike me at all...but it works. What I did before was somehting like this:
x² - x < 0
therefore x(x-1) <0
therefore x-1<0
therefore x < 1
But x-1<0,
Therefore x lies in the range 1 and 0.

Is this right??

Uhmm, not quite correct. Let f(x) = x² - x. And you want to solve for f(x) < 0, right?
If f(x) is continuous, and changes from positive to negative, or from negative to positive at some x₀ value, then at that x₀ value, (i.e the x₀ value that f(x) starts to change sign), f(x₀) = 0, right? So first, we solve for f(x) = 0, we have 2 value, 0, and 1.
So you'll have 3 intervals, we can then test to see the sign of f(x) in each interval:
$$\begin{tabular}{c|lrrr} x & - \infty & 0 & 1 & + \infty \\ \hline f(x) & & + \ \ \ \ \ 0 & \ \ \ - \ \ \ 0 & \ \ \ \ \ + \ \ \ \ \ \end{tabular}$$
So on the interval (0, 1), f(x) will take negative sign, i.e f(x) < 0.

 

[Hootenanny]

You may want to check your working there Viet, I have the range as 0<x<1.

Congrats on the HH medal btw

 

[Office_Shredder]

$$\begin{tabular}{c|lrrr} x & - \infty & 0 & 1 & + \infty \\ \hline f(x) & & + \ \ \ \ \ 0 & \ \ \ - \ \ \ 0 & \ \ \ \ \ + \ \ \ \ \ \end{tabular}$$

He even got that result:

2 value, 0, and 1.

I think he just went braindead for a second. It happens

 

[VietDao29]

You may want to check your working there Viet, I have the range as 0<x<1.

Whoops, a serious mistype. Should be more careful next time...

Congrats on the HH medal btw

Yah, thanks.

 

[MindRafter]

But the question was nothing related to functions...it's just from "Real Numbers and the Real Line", a preliminary of Calculus...
But whatever that may be - I guess my calculation wasn't very correct...
:(

Thanks :) ;) Viet, Hootenanny

 

[HallsofIvy]

But the question was nothing related to functions...it's just from "Real Numbers and the Real Line", a preliminary of Calculus...
But whatever that may be - I guess my calculation wasn't very correct...
:(

Thanks :) ;) Viet, Hootenanny

And why in the world would calculus not be related to functions?

Actually, the fact that this function, y= x²- x, must have the same sign on each of those intervals and can only change sign at x= 0 and x= 1, where the y= 0, is due to a fundamental property of any continuous function: If f(x₀)= a and f(x₁)= b, then f must take on every value between a and b between x₀ and x₁. If f is positive at one value of x and negative at another, it must take be 0 some place between.

You said earlier

therefore x(x-1) <0 therefore x-1<0

which is, of course, wrong. Dividing both sides of the inequality by x will give either x-1< 0 or x- 1> 0 depending on whether x itself is positive or negative.
You could have argued like this: x(x-1)< 0 so x and x- 1 must have different signs. If x< 0, then both x= x- 0 and x- 1 are negative (smaller number minus larger is negative). Their product is positive. If 0< x< 1, x= x-0 is positive while x-1 is still negative. Their product is negative. Finally, if x> 1 then both x and x-1 are positive and so their product is positive.

Yet a third way, for this simple problem, is to recognize that the graph of y= x²- x is a parabola, crossing the x-axis at x= 0 and x= 1 and opening upward. The middle part of the parabola is below the x-axis so x²- x< 0 for 0< x< 1.

 

[david1701]

Why the tables we were taught to let equalities be equals signs and there were no problems unless u divided by an unknown (which is do-able but not neccasery here) so:

[tex]x^2-x<0[\tex]

[tex]x^2-x=0[\tex]

x(x-1)=o

then x=1 and x=0

this means x<1 and x>0

so 0<x<1
x is between 0 and 1

why then need for y's and tables
why do people feel the need to do things the hard way, i have a physics teacher that does this, he will use "acurate" drawing over the cosine rule it's bizzare

 

[Hurkyl]

x(x-1) <0
therefore x-1<0

This is only a valid step if you already know that x is positive. You do not yet know that. So, if you want to do what you are doing, you have to divide the problem into three cases:

(1) x is positive
(2) x is zero
(3) x is negative

 

[turdferguson]

Why the tables we were taught to let equalities be equals signs and there were no problems unless u divided by an unknown (which is do-able but not neccasery here) so:

[tex]x^2-x<0[\tex]

[tex]x^2-x=0[\tex]

x(x-1)=o

then x=1 and x=0

this means x<1 and x>0

so 0<x<1
x is between 0 and 1

why then need for y's and tables
why do people feel the need to do things the hard way, i have a physics teacher that does this, he will use "acurate" drawing over the cosine rule it's bizzare

You need to use a sign graph because you don't know if the interval from 0 to 1 is positive or negative. If the question asked x²-x > 0 , your method of "just set them equal" would come up with the same interval. The correct answer is everything but 0 to 1, negative infinity to 0 or 1 to positive infinity.

Setting the function equal to zero and finding the roots establishes where the graph will change signs. But after that, you still need to see what the question asks for and plug in points to see which intervals work and which dont. As a rule of thumb for simple inequalities, the signs alternate at each root. But finding the roots is only half the problem