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# c3 Modulus watch

1. |x^2 +3x|=4x + 6

I got x = -1 or -6
and x= -2 or 3

the answer is -1 and 3.
But without looking at the graph, is there any way of getting rid of -6 and -2 ?
2. Yes.

Start by factorising to get . This is positive when or , and negative when .

So, we solve in the range , and we solve in the range .

In each case, you get two roots, but only one of the roots lies in the range, so you discard the other root.
3. Thanks, im guessing thats not required at c3 though? as iv done most test papers and its only come up twice, and both times, it was a 2 mark question.
Having to do that and find x would be worth a lot more marks im guessing.
4. (Original post by uxa595)
Thanks, im guessing thats not required at c3 though? as iv done most test papers and its only come up twice, and both times, it was a 2 mark question.
Having to do that and find x would be worth a lot more marks im guessing.
What do you mean? You're meant to know how to solve equations involving moduli, and these equations require you to consider different cases based on whether what's inside the modulus brackets is positive or negative. Another way to do it would be to draw the two graphs corresponding to and and work out the roots from there.
5. (Original post by uxa595)
|x^2 +3x|=4x + 6

I got x = -1 or -6
and x= -2 or 3

the answer is -1 and 3.
But without looking at the graph, is there any way of getting rid of -6 and -2 ?
I use a longer method but it works all the time

|x^2+3x|=4x +6
square both sides and solves - it does work I used it and have got the answers correct; yes it is longer but that's how I learnt.
6. (Original post by nuodai)
Yes.

Start by factorising to get . This is positive when or , and negative when .

So, we solve in the range , and we solve in the range .

In each case, you get two roots, but only one of the roots lies in the range, so you discard the other root.

I understand how that but can not just draw a graph for |x^2 +3x|=4x + 6
Which will then show my the ranges?

As i dont quite understand how you would have got the ranges for something like |x^2 -1|=6x using your method
7. I know its negative between -1<x<1 by just looking at it without making a graph but for a more complex more, would it not be easier just to draw a quick graph or paper or your calculator and see.
8. (Original post by uxa595)
I understand how that but can not just draw a graph for |x^2 +3x|=4x + 6
Which will then show my the ranges?

As i dont quite understand how you would have got the ranges for something like |x^2 -1|=6x using your method
Factorising gives . This is positive in the region and , and negative in the region , and so:

(i) In the region , you solve
(ii) In the region , you solve

Then you discard any roots not in the appropriate range of values of .

Can you see how you might apply this to other problems now?

(Original post by Nino9)
I use a longer method but it works all the time

|x^2+3x|=4x +6
square both sides and solves - it does work I used it and have got the answers correct; yes it is longer but that's how I learnt.
That would require you to be able to factorise quartics, and on top of that squaring sometimes introduces roots which aren't valid in the original equation. Squaring both sides of problems involving moduli really isn't a good idea unless you know for certain that it won't mangle your results (which it sometimes can).
9. (Original post by uxa595)
I know its negative between -1<x<1 by just looking at it without making a graph but for a more complex more, would it not be easier just to draw a quick graph or paper or your calculator and see.
Sketching the curves on the same graph is a good enough method for C3. Nuodai's method is "stronger" but it's not necessary to do it that way for C3 questions.
10. (Original post by uxa595)
|x^2 +3x|=4x + 6

I got x = -1 or -6
and x= -2 or 3

the answer is -1 and 3.
But without looking at the graph, is there any way of getting rid of -6 and -2 ?
Yeah, plug x = -6 and x = -2 into the original equation and notice the two sides aren't equal, hence you can discount them as solutions.

Probably wouldn't get full marks though, so just do a stupid sketch for the sake of the examiner and then ignore it and do it your way, making sure to check at the end that all your solutions work (and if some don't, then just write a comment like "as visible from the graph, x = -6 is not a solution) - that will get you full marks.

Personally, I think actually using a sketch properly is the best way to do it.

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Updated: December 24, 2010
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