Simple maths problem (Greater than/Less than) help

x(squared) - 7x - 18 > 0

Right I know what to do, you factorise the quadratic etc.

So you get (x + 2)(x - 9) > 0

I though -2<x<9 but apparently I'm wrong? I always get confused when it comes to using GREATER THAN/ LESS THAN signs and having to find a set of values which satisfy the equation. When are the x values greater than or greater and equal etc. ?

I'm sorry if I may not be explaining my problem very well, a simple question which commonly occurs in our AS-Level Maths papers but just don't know how to write the set of x values which satisfy the equation, when is it greater than or less than etc.

Thanks.

The Attempt at a Solution

Hurkyl
Staff Emeritus
Gold Member
Draw a picture. (Seriously)

As x varies, continuous functions cannot change their sign without passing through zero. If there are few zeroes, then that reduces the problem to just a handful of things to check via direct calculation. (and even less if you really know the behavior of the function in question)

jgens
Gold Member
For the equation to be greater than zero, what must be true of both of the factors?

Homework Helper
You can think of x^2 - 7x - 18 as a function, let's say f(x) = x^2 - 7x - 18. Can you plot the graph of this function, and see where f(x) > 0 holds? This should help.

D H
Staff Emeritus
So you get (x + 2)(x - 9) > 0
Correct.

I though -2<x<9 but apparently I'm wrong?
Correct (the "apparently I'm wrong" part, that is). Drawing a picture, as Hurkyl and radao suggested, will demonstrate this. Use your graphing calculator, if you have one.

Both x+2 and x-9 must have the same sign for (x+2)(x-9) to be positive. So, two cases to analyze: x+2 and x-9 are both negative, or x+2 and x-9 are both positive. Can you take it from here?

HallsofIvy
Homework Helper
Another way of looking at it is this: you know that (x+2)(x-9) is equal to 0 when x= -2 or x= 9. Those are the points where (x+2)(x-9) can change sign. Divide the real line into three parts, x< -2, -2< x< 9, and 9< x. Choose one value of x in each interval and check the sign at that point. For example, -3< -2 and (-3)2- 7(-3)- 18= 9+ 21- 18= 12> 0. Since the sign can change only at -2 and 9, x2- 7x - 18> 0 for all x< -2. -2< 0< 9 and (0)2- 7(0)- 18= -18< 0 so x2- 7x- 18< 0 for all x between -2 and 9. Finally, 10> 9 and (10)2- 7(10)- 18= 100- 70- 18= 12> 0. x2- 7x- 18> 0 for all x> 9.

I think graph technique is the best but I wasn't allowed to use that in the exam. My professor would use different cases .. solve each of them and in the end, combine everything using AND/OR symbols(a lot of mess and I never really remembered them). So, here's what I used:

(x+2).(x-9) > 0
WHEN
[(x+2) > 0 AND (x-9) > 0] OR [(x+2) < 0 AND (x-9) < 0]
note the distinction between AND, OR they are boolean
Now, it's an easier problem to solve. You just need to take care of AND's OR's. Instead of introducing AND/OR in the end, I state them in the beginning.

For quadratic functions of x that has solutions to determine which symbols to use depends on the co-efficient of x2.

A negative co-efficient will show an upside down U-shape and positive will show a U-shape for your curve. So, naturally if it has real roots then f(x) < 0 will be continous for positive co-efficients and broken for negative co-efficients and vica versa. It is a good idea to draw a graph or something because losing marks in an exam for questions like these is really annoying because it is something you can do quite easily.

Example

x2 -7x - 18
(x + 2)(x - 9)>0
x+2=0
x=-2
x-9=0
x=9
Since this is for > 0 it is a broken inequality.
x>9
x<-2
(x + 2)(x - 9)<0
Since this is for < 0 it is continous for a positive U shaped curve.
-2<x<9

For negative co-efficents it will be the opposite. Where (-x-2)(x-9)<0

x<9
x>-2
(-x-2)(x-9)>0
-2<x<9

EDIT: don't try to solve for (x+a)(x+b)>0
x>-a
b>-b

doesn't really work.

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