Modulus?

I have my Core 3 exam on Friday.

When you have modulus of two lines, I always find it really hard to draw them to find where they intersect. Other than trying lots of different combinations algebraically (which is fine and I can do- but if they are quadratic and one is a straight line you can get the wrong answers and it isn't always that reliable) is there any way you can find out where the two lines cross that you know will definitely get you the right answer?!

I have spoken to loads of people about it and no one has been able to provide a solution (not even my maths teachers!)

Please please help- sorry its rambled but am stressed! Thank you !

Whenever I see a modulus sign, my first instinct is to square it. This gets rid of the modulus sign and makes things much more manageable. Perhaps give an example of a question?

This is from AQA core 3 June 2011:

Solve the equation |3x+3|=|x^2-1|


I approached it by doing....

3x+3=x^2-1 and solving that
then
-3x-3=x^2-1

then -3x-3=-x^2 +1

and 3x+3=-x^2 +1


But it gave three answers when you solve all of the equations. How would you work it out?

Thankyou

Whenever I see a modulus sign, my first instinct is to square it. This gets rid of the modulus sign and makes things much more manageable. Perhaps give an example of a question?

squaring is not always correct ...

For the solution of equations it's fine. You then have to check the values obtained with the original equation/inequality.

exactly the point ... it is just that the checking bit was not mentioned... or maybe I missed it

This is how I would do it, by no means is this the only way to do it! My first step would be to square both sides:

$|3x+3| = |x^2-1|$
$(3x+3)^2 = (x^2-1)^2$

Expand the brackets, simplify, and factorise.

EDIT: Notice that $(3x+3)^2 = 9(x+1)^2$ and $x^2-1 = (x+1)(x-1)$

It wasn't mentioned, my mistake

exactly the point ... it is just that the checking bit was not mentioned... or maybe I missed it

It wasn't mentioned, my mistake

Would you get the same answers using that method? And how would you check that they were correct? Would it be a pain to work it through step by step? Thanks

Whats the checking thing?!

You would get the same answers We aren't supposed to go through full solutions, read my edit to make the factorising much easier. Have you tried it?



The checking thing is required because sometimes solutions can be introduced by squaring, so for each value you obtain you need to plug this into the original equation (with the modulus signs) to check that this value satisfies it.

How would you do the same question then?

Surely you would end up with a horrible equation in X^4? How would you then solve that? Ahhh i don't like Maths!

either by graphing first


or

|f(x)| =|g(x)|

then

f(x) = g(x)
f(x) =- g(x)




Rules of thumb
|f(x)| =|g(x)|

then

f(x) = g(x)
f(x) =- g(x)



|f(x)| =c

then

f(x) = c
f(x) =- c




|f(x)| =g(x)

then

f(x) = g(x)
f(x) =- g(x)
then you must check against original




|f(x)| =|g(x)|+h(x)

then

|f(x)| =g(x)+h(x)
|f(x)| =-g(x)+h(x)
then you must check against original

Is that a similar way to how I solved it originally then? I think this was the method that we were taught? It may be a stupid question, but how do you check against the original?

My advice is to stick to the method you were shown at school and if you are in any doubt check at the end by substituting your values into the original modulus equation.
I do not think you need to worry too much