Maths modulus question help please

13

Hi, please could I have some help on this question. I was attempting this modulus question and I'm stuck on part c. To find the point of intersection, I did 8- (2t-6)^2 = (t-3)^2+ 4 but I got the incorrect answer. I also saw a video, I understand how it is intersecting the positive and negative gradient but then I'm unsure how they got the equations like that?
Thank you!

Reply 1

anonymous56754

OP

13

Original post

by anonymous56754

Hi, please could I have some help on this question. I was attempting this modulus question and I'm stuck on part c. To find the point of intersection, I did 8- (2t-6)^2 = (t-3)^2+ 4 but I got the incorrect answer. I also saw a video, I understand how it is intersecting the positive and negative gradient but then I'm unsure how they got the equations like that?
Thank you!

Reply 2

mqb2766

21

Original post

by anonymous56754

Hi, please could I have some help on this question. I was attempting this modulus question and I'm stuck on part c. To find the point of intersection, I did 8- (2t-6)^2 = (t-3)^2+ 4 but I got the incorrect answer. I also saw a video, I understand how it is intersecting the positive and negative gradient but then I'm unsure how they got the equations like that?
Thank you!

You cant just square up part of an expression in an equation. If you squared both sides of an equation, so keep balance, then thats ok but here its not an easy way to go.

The usual mod way, which you should practice/know, is to solve on each interval where you can remove the mod sign. So here the domains are t<=3 and t>=3 and these work for both mod expressions so for the domain t<=3 then
-(t-3) + 4 = 8 - -(2t-6)
as both mod args are negative on this domain ... A solution for t is only valid if its in the original domain so <=3 here.

Then the opp for t>=3.

Reply 3

anonymous56754

OP

13

Original post

by mqb2766

You cant just square up part of an expression in an equation. If you squared both sides of an equation, so keep balance, then thats ok but here its not an easy way to go.
The usual mod way, which you should practice/know, is to solve on each interval where you can remove the mod sign. So here the domains are t<=3 and t>=3 and these work for both mod expressions so for the domain t<=3 then
-(t-3) + 4 = 8 - -(2t-6)
as both mod args are negative on this domain ... A solution for t is only valid if its in the original domain so <=3 here.
Then the opp for t>=3.

Ohh ok thank you! Just to clarify, when there are two mods, we take negative of both and solve, then the positive of both?
Also what do you mean by the domain being t<=3 and t>=3 , doesn't the question say the domain is t>=0?

Reply 4

mqb2766

21

Original post

by anonymous56754

Ohh ok thank you! Just to clarify, when there are two mods, we take negative of both and solve, then the positive of both?
Also what do you mean by the domain being t<=3 and t>=3 , doesn't the question say the domain is t>=0?

You should be able to look this up but the two mod functions are
|t-3|
|2t-6|
and the arguments t-3 and 2t-6 are both zero at t=3, so thats the point where the gradient jumps/the vertex of the "v" or "^" in the mod function. Either side of that you can replace the mod function with either
+/-(t-3)
+/-(2t-6)
as appropriate. The key idea is to solve two equations independently instead of one, where the two equations are only defined on part of the original domain. However, for those two domains/intervals you can replace the mod function with a linear function (in this case) so each equation is easy to solve, on its restricted domain.

So consider the domain//intervial t<=3. As you say there is an additional constraint/inequality t>=0, so really its 0<=t<=3. As both t-3 and 2t-6 are negative on this domain (sub t=0 for instance), then the mod functions are equal to
|t-3| = -(t-3)
|2t-6| = -(2t-6)
so simply replace the mod expressions with the two right hand sides and solve the equation on this domain.

Similarly for the domain t>=3 which you should work through.

Reply 5

anonymous56754

OP

13

Original post

by mqb2766

You should be able to look this up but the two mod functions are
|t-3|
|2t-6|
and the arguments t-3 and 2t-6 are both zero at t=3, so thats the point where the gradient jumps/the vertex of the "v" or "^" in the mod function. Either side of that you can replace the mod function with either
+/-(t-3)
+/-(2t-6)
as appropriate. The key idea is to solve two equations independently instead of one, where the two equations are only defined on part of the original domain. However, for those two domains/intervals you can replace the mod function with a linear function (in this case) so each equation is easy to solve, on its restricted domain.
So consider the domain//intervial t<=3. As you say there is an additional constraint/inequality t>=0, so really its 0<=t<=3. As both t-3 and 2t-6 are negative on this domain (sub t=0 for instance), then the mod functions are equal to
|t-3| = -(t-3)
|2t-6| = -(2t-6)
so simply replace the mod expressions with the two right hand sides and solve the equation on this domain.
Similarly for the domain t>=3 which you should work through.

i see, but what if the domains are not given, how would we be able to tell whether the mods are negative or positive? Or do we just square in that instance?
also did you get the domain t=>3 from t-3=0?

Reply 6

mqb2766

21

Original post

by anonymous56754

i see, but what if the domains are not given, how would we be able to tell whether the mods are negative or positive? Or do we just square in that instance?
also did you get the domain t=>3 from t-3=0?

The domain t>=0 in the question is fairly irrelevant for this discussion which is why I didnt refer to it in the original reply.

For the mod function (again, pls look this up), the mod/abs value function is defined by
|x| = x if x>=0
|x| = -x if x<=0
So there is a gradient jump of -1 to 1 at x=0. Here just consider the |t-3| mod functtion so the argument t-3=0 at t=3. This is where the mod functions graident jumps so
|t-3| = t-3 if t>=3
|t-3| = -(t-3) if t<=3
...

Similarly for |2t-6| as its equal to |2(t-3)| = 2|t-3|, so ...

So you decide to split the problem at t=3, based on what the arguments of the mod functions are. If there were two mod functions and they "jumped" (args are zero) at different places (values of t), then youd analyse 3 seperate problems, each with a restricted domain defined by where they "jumped".

With regard to the op, I guess youve squared stuff up in the past? Thats ok for simple mod equations involving a single mod function (as long as you remember to reject the extraneous solutions). However, for more complex cases it can cause more problems than doing it the right/usual way as above.

(edited 8 months ago)

Maths modulus question help please

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