Laws of inequalities

Hey,
So I'm studying to resit my AS maths C1&2 modules.
When practicing solving inequalities, I've come across something that confused me quite a lot, any help is appreciated

So, the sign for an inequality changes direction when we multiply/divide both sides by a negative number. However, after attempting practice questions I have noticed that the book disagreed with my answer, and the only way to reach the correct one was to swap the direction of the inequality sign whenever we add a number to both sides.

Consider (3x^2-13x+4<0)

This quadratic inequality is factorised to:
(3x-1)(x-4)<0

Therefore
x-4<0
or
3x-1<0

x-4<0
x<4

3x-1<0
3x<1
x<1/3

So the overall solution that I come to is x<1/3 as it appears to satisfy both solutions above, but the book insists that the overall solution is:
1/3<x<4

and I'm melted trying to figure out how to reach that using the rules for flipping the inequalities that I've learned, maybe I'm missing something. If anyone could explain this to me I'd be very grateful

My Thanks,
PFC Algeo UKCM

Hey,
So I'm studying to resit my AS maths C1&2 modules.
When practicing solving inequalities, I've come across something that confused me quite a lot, any help is appreciated

Hey,
So I'm studying to resit my AS maths C1&2 modules.
When practicing solving inequalities, I've come across something that confused me quite a lot, any help is appreciated

So, the sign for an inequality changes direction when we multiply/divide both sides by a negative number. However, after attempting practice questions I have noticed that the book disagreed with my answer, and the only way to reach the correct one was to swap the direction of the inequality sign whenever we add a number to both sides.

Consider (3x^2-13x+4<0)

This quadratic inequality is factorised to:
(3x-1)(x-4)<0

Therefore
x-4<0
or
3x-1<0

First of all, thankyou for replying!
Abby I understand what you mean, and that's helped me a great deal, I suppose now I'm confused as to how I could work through that problem without drawing the graph, which I don't need to know.. I would just like as full an understanding as possible, but I feel like I can go out and tackle these questions now anyhow

Davros I'm not sure what you mean; how are we to know that (3x-1>0)? Why couldn't it be 3x-1<0? That's all I'm struggling to understand at this point; but I think I can work out these questions now.

First of all, thankyou for replying!
Abby I understand what you mean, and that's helped me a great deal, I suppose now I'm confused as to how I could work through that problem without drawing the graph, which I don't need to know.. I would just like as full an understanding as possible, but I feel like I can go out and tackle these questions now anyhow

Davros I'm not sure what you mean; how are we to know that (3x-1>0)? Why couldn't it be 3x-1<0? That's all I'm struggling to understand at this point; but I think I can work out these questions now.

First of all, thankyou for replying!
Abby I understand what you mean, and that's helped me a great deal, I suppose now I'm confused as to how I could work through that problem without drawing the graph, which I don't need to know.. I would just like as full an understanding as possible, but I feel like I can go out and tackle these questions now anyhow

Davros I'm not sure what you mean; how are we to know that (3x-1>0)? Why couldn't it be 3x-1<0? That's all I'm struggling to understand at this point; but I think I can work out these questions now.

If your quadratic is negative, what can you say about the two expressions as individuals?

If the coefficient of the x^2 term is positive then the curve has a minimum point, negative then its a maximum point, then the direction of the inequality sign will tell you which part of the graph (above/below x axis) holds your answer, or that's my understanding anyway

You need to quote me if you want me to see that you've replied to me

If you multiply two numbers A and B to get a negative number then one of the original numbers must be positive and the other negative. So the two cases you need to consider are:

A > 0 AND B < 0

or

A < 0 AND B > 0

I suspect if you follow your original strategy but take account of the conditions on BOTH brackets then you will arrive at the correct answer.

The graphical method is also sound - if you find the points where the curve cuts the axis, then just choose 3 other points to plot - one to the left of the 1st intersection, one between the two intersections, and one to the right of the 2nd intersection. This will show you whether the curve is above or below the axis in each region.