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  1. #1
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    Question

    (i)Sketch on the same diagram the graphs of y = |2x+3| and y = 1-x
    (ii)Find the values of x for which x + |2x+3|= 1
    Last edited by Matt512; August 2nd 2013 at 07:00 PM.
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  2. #2
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    Re: Question

    Are you using the /symbol to represent absolute values? If so, you should really use |
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  3. #3
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    Re: Question

    I equated the two equation and got x^3 - 5x + 2= 0
    I find the f(2) = 0 which is (x-2) is a factor of x^3 - 5x + 2
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  4. #4
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    Re: Question

    Hi matt512,
    The attachment shows the graph and indicates how you can solve the equation.

    Question-mhfgeometry34.png
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  5. #5
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    Re: Question

    Quote Originally Posted by Matt512 View Post
    I equated the two equation and got x^3 - 5x + 2= 0
    I find the f(2) = 0 which is (x-2) is a factor of x^3 - 5x + 2
    I have no idea what you are saying here. HOW did you "equate" the two equations? One is linear and the other an absolute value. I can't imagine how you could get a cubic from that. And then you say "I find the f(2)". What??? There is no "f" anywhere in the problem!

    From the definition of "absolute value", |2x+ 3| is either 2x+ 3 or -(2x+ 3) depending upon whether 2x+3 is positive or negative. If 2x+3\ge 0, then |2x+3|= 2x+ 3. Setting that equal to 1- x gives 2x+ 3= 1- x. Adding x- 3 to both sides, 3x= -2 so x= -2/3. Checking, 2(-2/3)+ 3=-4/3+ 9/3= 5/3 which is, indeed positive so that is a valid solution. If 2x+ 3\le 0, then |2x- 3|= -(2x- 3)= 3- 2x. Setting that equal to 1- x give 3- 2x= 1- x. Adding 2x- 1 to each side 4= x. 2(4)+ 3= 11 which is positive so that is a valid solution.

    |2x+ 3|= 1- x is equivalent to x+ |2x+ 3|= 1 (add x to both sides). They have the same solutions.
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