where is the second solution?

A

allegansveritatem

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I worked out this problem in what seems a logical manner, but only got one solution. What is it that I am not doing correctly?
Here is what I did:
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Well which is it, x<9/25 or x<3/5 or are the same?

You got the wrong answer.

You were not careful making sure that you were not dividing by a negative (or unknown) value which would( might) require you to change the direction of the inequality sign.

Also were you careful when you took the square root of both sides? For example, we know that -4< 25 but \(\displaystyle \sqrt{-4}\)\(\displaystyle \nleq\)\(\displaystyle \sqrt{25}\)

Personally I would simply factor. Then you have a product of two factors that must be less than 0. So one is positive while the other is negative AND visa versa.

I do not mind your asking us to check your work but you should check it yourself. Imagine if x is very negative (large enough to make my point true), then 5x^2 will be much much larger than 9x making 5x^2-9x > 0.

One clear mistake is when you took the square root of x, that MUST mean that x is not negative. Yet your answer for x allows ALL negative numbers.

My advise is to not take square roots with inequalities unless you are extremely careful and do not divide by unknowns unless you know they are always positive or always negative OR you break the problem into parts.
 
Last edited:
allegansveritatem said:
I worked out this problem in what seems a logical manner, but only got one solution. What is it that I am not doing correctly?
Here is what I did:
View attachment 11789
Click to expand...
The way I do these is to get everything on one side of the inequality. I then look for "critical points," points at which the function is zero, infinite, or you are doing division by zero. Then you can set up regions where the inequality works or does not work.

In your example you have \(\displaystyle 25x^2 - 9x < 0\)

As Jomo suggested we factor the LHS:
\(\displaystyle x(25x - 9) < 0\)

This has two critical points: x = 0 and x = 9/25. So split up the number line and we have three possibilities for where the inequality is correct: \(\displaystyle ( - \infty, 0 ] \text{, and } (0, 9/25 ] \text{, and }(9/25, \infty )\)

By choosing numbers from each interval we can get that the answer is (0, 9/25).

If this seems not to be enough "analytical" then consider we have to find the intervals anyway, else we can't decide where x has to be. (The negative signs and all that.)

-Dan
 
Jomo said:
Well which is it, x<9/25 or x<3/5 or are the same?

You got the wrong answer.

You were not careful making sure that you were not dividing by a negative (or unknown) value which would( might) require you to change the direction of the inequality sign.

Also were you careful when you took the square root of both sides? For example, we know that -4< 25 but \(\displaystyle \sqrt{-4}\)\(\displaystyle \nleq\)\(\displaystyle \sqrt{25}\)

Personally I would simply factor. Then you have a product of two factors that must be less than 0. So one is positive while the other is negative AND visa versa.

I do not mind your asking us to check your work but you should check it yourself. Imagine if x is very negative (large enough to make my point true), then 5x^2 will be much much larger than 9x making 5x^2-9x > 0.

One clear mistake is when you took the square root of x, that MUST mean that x is not negative. Yet your answer for x allows ALL negative numbers.

My advise is to not take square roots with inequalities unless you are extremely careful and do not divide by unknowns unless you know they are always positive or always negative OR you break the problem into parts.
Click to expand...
Yes, I see what you are saying. I actually mad a mistake here when I put that negative infinity there and I don't even know now what I was tinking about. It should be (0, 9/25). The 3/5 was a mistake. I was thinking somehow that 9/25 reduced down to 3/5. I think what I was thinking was that the sqrt of 9/25 is 3/5. Of course, that is not what was being asked for here. Yeah, factoring is the better way in this case. Thanks
 
Thanks for the "like", MarkFL. It made me look at my method again. The list of possible intervals should be
\(\displaystyle (- \infty, 0) \text{, and } (0, 9/25) \text{, and }(9/25, \infty)\)

We have to leave out x = 0 and x = 9/25 as they make the LHS 0, which is not allowed by the inequality.

-Dan
 
topsquark said:
The way I do these is to get everything on one side of the inequality. I then look for "critical points," points at which the function is zero, infinite, or you are doing division by zero. Then you can set up regions where the inequality works or does not work.

In your example you have \(\displaystyle 25x^2 - 9x < 0\)

As Jomo suggested we factor the LHS:
\(\displaystyle x(25x - 9) < 0\)

This has two critical points: x = 0 and x = 9/25. So split up the number line and we have three possibilities for where the inequality is correct: \(\displaystyle ( - \infty, 0 ] \text{, and } (0, 9/25 ] \text{, and }(9/25, \infty )\)

By choosing numbers from each interval we can get that the answer is (0, 9/25).

If this seems not to be enough "analytical" then consider we have to find the intervals anyway, else we can't decide where x has to be. (The negative signs and all that.)

-Dan
Click to expand...
.
I'm too baroque minded to come upon this nice simple method. I think what endeared me to the way I tried to do it was the fact that I could use one of my "discoveries" namely that a variable divided by its square root is the square root of said variable. Wow! Thanks for posting
 
allegansveritatem said:
I worked out this problem in what seems a logical manner, but only got one solution. What is it that I am not doing correctly?
Click to expand...
The problem is to solve \(\displaystyle 25x^2-9x<0\).
\(\displaystyle \begin{align*}25x^2-9x&<0 \\x(25x-9)&<0 \end{align*}\)

Hence there are two critical values: \(\displaystyle x=0~\&~x=\tfrac{9}{25}\)
Those two values create three areas: \(\displaystyle I(-\infty,0),~II\left(0,\tfrac{9}{25}\right)~\&~III\left(\tfrac{9}{25},\infty\right)\)
We will test each of the three:
\(\displaystyle \begin{align*}x&=-1\in I\text{ FALSE} \\x&=0.1\in II\text{ TRUE}\\x&=1\in III\text{ FALSE} \end{align*}\)

Thus now we know that the inequality holds only on area \(\displaystyle II\).

 
Jomo said:
By the way, what did you mean by Where is the 2nd solution??
Click to expand...
I think the problem is that I am not grasping the concept of inequality. It is fuzzy. I am treating these problems as though they were solvable in the same way and with the same scope as equalities. Thus, since this is a quadratic expression, I was thinking that it should have two solutions. But what is wanted here is something else, something like a range of values, rather than a pair of them. I have been familiar with this notion of inequality for a long time but now it is becoming more complex and I have to expand my understanding.
 
pka said:
The problem is to solve \(\displaystyle 25x^2-9x<0\).
\(\displaystyle \begin{align*}25x^2-9x&<0 \\x(25x-9)&<0 \end{align*}\)

Hence there are two critical values: \(\displaystyle x=0~\&~x=\tfrac{9}{25}\)
Those two values create three areas: \(\displaystyle I(-\infty,0),~II\left(0,\tfrac{9}{25}\right)~\&~III\left(\tfrac{9}{25},\infty\right)\)
We will test each of the three:
\(\displaystyle \begin{align*}x&=-1\in I\text{ FALSE} \\x&=0.1\in II\text{ TRUE}\\x&=1\in III\text{ FALSE} \end{align*}\)

Thus now we know that the inequality holds only on area \(\displaystyle II\).

Click to expand...
Good. I am getting it. Thanks
 
25x^2 - 9x < 0

In my opinion, anything that says to "flip the inequality" or "reverse the inequality" is misguided and inherently confusing. Just don't do it.

Solve this problem: 25x^2 - 9x = 0. You should find two solutions. These two solutions divide the number line into three distinct parts. In which of these parts is the result negative?

Solve this problem: x(25x-9) < 0. When multiplying two numbers, they must have opposite signs if the result is to be negative. So, ask 2 questions: 1) Is 25x-9 > 0 when x < 0, and 2) Is 25x-9 < 0 when x > 0

No need to "flip" anything.

Also, please do not do things that are not reversible.

It is true that -4 < 4.
It is not true that (-4)^2 < 4^2
When faced with 16 < 16, can you derive from this that -4 < 4? It should not make sense.
So, when faced with 25x^2 < 9x, do you REALLY think it appropriate simply to apply a square root to both sides?
 
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