Maths A level Question

Hi,

What is your answer to this question....

Find the set of values of x for which: X² - 9X < 36

I found it to be..

X² - 9X - 36 < 0
X² +3X - 12X - 36 < 0
X(X+3)-12(X+3) < 0
(X+3)(X-12) < 0
X=-3 X=12
-3<X<12

Many thanks

x

You've missed an answer, probably because you didn't understand what you did when going from (x+3)(x-12)<0 to -3<x<12. (This is why it's good to get into the habit of explaining your work as you go along)

Note that ab<0 implies a,b have different signs, i.e. a<0 and b>0 or a>0 and b<0. So with a=x+3 and b=x-12 we have either:
1) x+3<0 and x-12>0 which leads to x<-3 and x>12.
2) x+3>0 and x-12<0 which leads to your answer.

No, the function is greater than zero for values of x less than -3 or greater than 12 so your first answer is incorrect. The OP did it right in the first place.

I stand corrected. Wasn't concentrating there, sorry. The reason it doesn't work is because they can't hold simultaneously as I erroneously assumed at first (hence why I said "x<-3 AND x>12", which is absurd), so you won't have x+3,x-12 having different signs and so (x+3)(x-12)>0.

I stand corrected. Wasn't concentrating there, sorry. The reason it doesn't work is because they can't hold simultaneously as I erroneously assumed at first (hence why I said "x<-3 AND x>12", which is absurd), so you won't have x+3,x-12 having different signs and so (x+3)(x-12)>0.

You was correct, because you solved the inequality algebrically using logical statments
(a>0, b<0 ....) and logical functions (AND, OR)
As you wrote there is 2 cases
In every case you have to solve an inequality system simoultaneously, Every
solution of an ineguality is a real interval. THe solution for the system with
2 inequalities will be the intersection of the two real intervals (there is logical AND between the inequalities)
So. 1. x+3<0 AND x-12>0
x<-3 1st interval AND
x>12 2nd interval
The intersection is null set that is there is no solution here
OR
2. x+3>0 AND x-12<0
x>-3 1st interval (]-3; +infinity[) AND
x<12 2nd interval
The intersection is -3<x<12

The full solution is the UNION of two cases but there is solution in the 2nd case only, so
this willl be -3<x<12


THat is another method when using (drawing) graph you read down that
for what x value will be the function negative.

Hi,

To clarify what have i got wrong because i answered the question above exactly as shown however, i did not receive full marks. what am i missing?

Many thanks

Hi,

Out of four marks i lost two, 1 for method and 1 for answer. Im home taught (teaching myself A levels) so i have no teacher to ask.

Many thanks

S

Hi,

Out of four marks i lost two, 1 for method and 1 for answer. Im home taught (teaching myself A levels) so i have no teacher to ask.

Many thanks

S

I thought it seemed harsh too, i have had a look back through five other exam questions with differing numbers which were exactly the same and they dont show anymore working than i have shown above..

I'm not familiar with A level mark schemes, but looking at what you have written there is no explanation of what you are doing - it's just a stream of symbols!

You certainly shouldn't write down X = -3, X = 12 immediately after an inequality because that isn't the correct answer - it's the answer to a different question which is "when does (x+3)(x-12) = 0?".

Also, as pointed out earlier, when you have an inequality of the form AB < 0, there are 2 possibilities that need to be considered:
A<0 and B >0
or
A>0 and B<0

In your case one of these leads to a contradiction and isn't valid, but you've not shown any working to indicate that you understand this!

How would you answer the question? just using numbers no need to reference back to the OP just what you would write if the question was placed in front of you.

How would you answer the question? just using numbers no need to reference back to the OP just what you would write if the question was placed in front of you.

Assuming what was given in the OP was the correct question (see comment from tinyhobbit above), I would lay it out like this:

$x^2 - 9x < 36 \Rightarrow x^2 - 9x - 36 < 0$

The LHS factorizes as (x - 12)(x + 3) so we have (x - 12)(x + 3) < 0.

There are 2 cases to consider:
(i) x -12 < 0 and x + 3 > 0
(ii) x - 12 > 0 and x + 3 < 0

For case (i) we need x < 12 and x > -3 which is OK.
For case (ii) we need x > 12 and x < -3 which is impossible.

Therefore the solution is -3 < x < 12

If you get into the habit of laying out your work like this then (a) it's much easier for you to read and check your thought processes when you check the answer at the end; (b) it's easier for other people (like an examiner) to see what you're doing

This is one particular topic in maths that bugs me, I can do Hard integration and 3D vectors but struggle to get my head around this.

Could you explain a little more why you have to consider two cases? is it todo with the graph could be u shaped or inverted?

Many thanks

S

You're just using the principle that if you multiply two numbers together and get an answer that is negative then one of the numbers must be negative and the other one positive. However, you don't know which number is negative and which one is positive! Therefore you have to consider both cases.

You can certainly answer the question graphically if you're confident that you can draw it accurately.