Modulus problems

Avva · 2009-10-09 02:50:20

FIND the solution set of the following equation:
X^2-3(modulus x-2)-4=0
Book's Ans:
If x≤2:x^2+3(x-2)-4=0
Therefore x^2+3x-10=0
Therefore x=-5 or 2 ]- ∞,2]
If x>2: x^2-3(x-2)-4=0
Therefore x^2-3x+2=0
therefore x=2 or 1 refused because they ]2, ∞]
S.S={-5,2}
on wt bases did he put negative signs in the solution ??!
Plzzzzz HELP
sorry i cant write the equation better than that

bobbym · 2009-10-09 12:31:28

Hi Avva;

I am having trouble following your work. What does X^2-3(modulus x-2)-4=0 mean?

Do you want to solve the equation modulo 2 or modulo x - 2. The whole equation or just (X^2 -3)? Does X = x ?

Last edited by bobbym (2009-10-09 12:54:47)

Avva · 2009-10-09 23:39:23

Sorry i couldn't write it well at first
x^2-3|x-2|-4=0

soroban · 2009-10-10 00:18:03

Hello, Avva!

. . .

.

Avva · 2009-10-10 03:53:54

Tx Bro U helped me really

Akh · 2012-07-22 17:58:05

Could you please help me in solving the below data sufficiency question.

If y > 0, what is the value of x?
(1) |x 3| > y
(2) |x 3| < y

Bob · 2012-07-22 20:05:36

hi Akh

Welcome to the forum.

I find it helps to have a graph for these absolute value questions.

Sketch y = x - 3 and then 'bend' the negative part round so it is positive. You end up with a V shaped graph. (see diagram below)

So, if y > 0, => |x-3| > 0

(1) Consider x > 3 and x < 3 separately.

(a) x > 3

problem becomes

x-3 > y > 0 => x > y + 3

(b) x < 3

problem becomes

3 - x > y > 0 => 3 > x + y

(2)

|x-3| < - y

(a) x>3

problem becomes

x-3 < -y => x + y < 3

(b) x<3

problem becomes

3-x < -y => y < x - 3

Hope that helps,

Bob

anonimnystefy · 2012-07-23 00:19:36

Hi Bob

I don't think your answer is correct. y isn't 0 it's less than or equal to it, so the answer should be in terms of y. e.g. Not all reals except 3 are answers to |x-3|>5.

Bob · 2012-07-23 03:11:12

Good point. I'm talking rubbish again.

I shall edit that post.

Thanks,

Bob

anonimnystefy · 2012-07-23 03:12:55

You are welcome. I'm a better proof-reader than I thought. Still not a good one, though.

Bob · 2012-07-23 03:22:36

You are excellent for keeping me from talking rubbish and that's a great help. I've edited the post now. Please cast your proof reading eye and tell me if you think I've got it now.

Bob

anonimnystefy · 2012-07-23 03:33:52

Hi Bob

(1) is okay now. I just recommend solving totally for x in terms of y. i.e. turn 3>x+y into x<3-y.

(2) is partially okay, but you will see that in the first case you have x>3 which comes out later to x+y<3 <=> x<3-y<3. So in that case we have no solution. Same for the other case in (2).

Bob · 2012-07-23 03:46:59

I showed a graph. I think I ought to use it more. I should be able to show shaded regions, some of which won't exist (??)

I'll post back when I've got some diagrams.

Bob

Bob · 2012-07-23 04:37:46

Ok, I have a satisfactory set of diagrams for this.

(1) x > 3 => x > y + 3 and x < 3 => 3 > x + y

I have shaded these two regions on my first diagram below.

(2) x > 3 => 3 > x + y and x < 3 => y < x - 3

I have shaded these on my second diagram but as we also know y > 0 my shaded regions are outside this constraint so, as anonimnystefy has said, no solutions exist.

Hopefully, that completes this problem.

Bob

Math Is Fun Forum

#1 2009-10-09 02:50:20

Modulus problems

#2 2009-10-09 12:31:28

Re: Modulus problems

#3 2009-10-09 23:39:23

Re: Modulus problems

#4 2009-10-10 00:18:03

Re: Modulus problems

#5 2009-10-10 03:53:54

Re: Modulus problems

#6 2012-07-22 17:58:05

Re: Modulus problems

#7 2012-07-22 20:05:36

Re: Modulus problems

#8 2012-07-23 00:19:36

Re: Modulus problems

#9 2012-07-23 03:11:12

Re: Modulus problems

#10 2012-07-23 03:12:55

Re: Modulus problems

#11 2012-07-23 03:22:36

Re: Modulus problems

#12 2012-07-23 03:33:52

Re: Modulus problems

#13 2012-07-23 03:46:59

Re: Modulus problems

#14 2012-07-23 04:37:46

Re: Modulus problems

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