C3 modulus functions

Hi everyone,

I have a question that I can't get the right answer for. It is:

solve gf(x)=|x^2 - 12| = x

I simply put it through the quadratic equation. I got x=-3 and x=4. But the book as it as x=3, x=4.
When I did the modulus of it (i.e. change the signs around) the discriminant was negative so I didn't go any further.

Have I just made a silly mistake or is there something else I did wrong...or is the book wrong?

Reply 1

james22

16

The LHS is always positive, so you should be able to see that x must always be positive. The book is right.

What was your working?

Reply 2

marcsaccount

OP

5

When you deduct x from both sides to make it equal zero, the x on the LHS is negative though? That's what I thought.

I had x^2-x-12 and used the quadratic formula.

Reply 3

marcsaccount

OP

5

I got

\frac{1 \pm 7}{2}

So -3 or 4.

Reply 4

natnalie

10

What you need to do is consider the graphs for both parts. The graph for y=x^2-12 is simply a smiley face that crosses the y-axis at the point (0,-12). But since its the modulus that you care about, the bottom half of the graph is reflected in the x-axis. So now the graph y=lx^2-12l crosses the y-axis at the point (0,12). Now you need to consider y=x. Its simply a straight line that crosses the origin. Therefore, combining these two graphs together gives you an intersection in the modulus of x^2-12. So its -(x^2-12)=x This means the quadratic is x^2+x-12=0. Which gives you x=-4 and x=3. Using the graph you can see that the intersection at x=-4 is reflected. Therefore this root is actually supposes to be x=4.

Reply 5

marcsaccount

OP

5

Original post by natnalie

What you need to do is consider the graphs for both parts. The graph for y=x^2-12 is simply a smiley face that crosses the y-axis at the point (0,-12). But since its the modulus that you care about, the bottom half of the graph is reflected in the x-axis. So now the graph y=lx^2-12l crosses the y-axis at the point (0,12). Now you need to consider y=x. Its simply a straight line that crosses the origin. Therefore, combining these two graphs together gives you an intersection in the modulus of x^2-12. So its -(x^2-12)=x This means the quadratic is x^2+x-12=0. Which gives you x=-4 and x=3. Using the graph you can see that the intersection at x=-4 is reflected. Therefore this root is actually supposes to be x=4.

Hi, thanks for your reply and explanation - it kinda makes sense :wink:

The only thing is that up until now I've been following the principal that when you make something equal zero, even with modulus equations, you still put in on the LHS and change the sign. I thought this was the correct method as my answers so far haven't been wrong.

So does what you're saying, that when we make it equal zero, it keeps it's same + or - sign even when we take it from one side to the other, does this just apply to this specific question or is it a general principal? I'm worried I'm really missing something!

Actually, the more I think about this the more I get confused. As it is modulus, if we reflect it in the line y=0, then x=-4 should still be x=-4, shouldn't it?

(edited 10 years ago)

Reply 6

marcsaccount

OP

5

This is what I mean, if you skip to 33min 50sec on this video https://www.youtube.com/watch?v=eh5HnB6Jg-s moving something from the RHS to the LHS, with a modulus equation, still requires the changing of the the +/- sign. So why don't we do it with the equation I posted above?

Reply 7

Merdan

10

I dont know about x=-4 transforming to x=4. But when you sketch y=modulus x^2-12 and y=x on the same diagram you will see that there are only 2 points where they cross which are occupied above the positive side of the x axis when i solve it x^2-12=x u get the 4 and when you solve -(x^2-12)=x u get ur 3 which is the second value what ur looking for

Reply 8

Merdan

10

Original post by Merdan

I dont know about x=-4 transforming to x=4. But when you sketch y=modulus x^2-12 and y=x on the same diagram you will see that there are only 2 points where they cross which are occupied above the positive side of the x axis when i solve it x^2-12=x u get the 4 and when you solve -(x^2-12)=x u get ur 3 which is the second value what ur looking for

Oh yes and the other two negative values -3 and -4 u know one of them would be your intersection value of two functions as if you didnt take the modulus of one function

Reply 9

marcsaccount

OP

5

Original post by Merdan

I dont know about x=-4 transforming to x=4. But when you sketch y=modulus x^2-12 and y=x on the same diagram you will see that there are only 2 points where they cross which are occupied above the positive side of the x axis when i solve it x^2-12=x u get the 4 and when you solve -(x^2-12)=x u get ur 3 which is the second value what ur looking for

Hi Merdan,

Thanks for your help, I understand it now and thankfully it falls in line with my previous understanding so no gaps :biggrin:

Reply 10

james22

16

Original post by marcsaccount

When you deduct x from both sides to make it equal zero, the x on the LHS is negative though? That's what I thought.

I had x^2-x-12 and used the quadratic formula.

If x was negative then you would have |stuff|=negative which is impossible because |stuff| is always non-negative.

C3 modulus functions

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