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    I don't understand why Meathod 1 is giving me the incorrect answer. It should technically give me the same answer as meathod 2??
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    you really need to show us the question :dontknow:
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    (Original post by FHL123)
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    ITs q18
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    i would start by plotting both curves on a graphics calculator to see roughly what is going on.
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    (Original post by the bear)
    i would start by plotting both curves on a graphics calculator to see roughly what is going on.
    Drawing the curves doesn't help
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    (Original post by FHL123)
    I don't understand why Meathod 1 is giving me the incorrect answer. It should technically give me the same answer as meathod 2??
    Actually both methods give the same answers. For method one, notice that you have  (3x-2)(2x+1)(2x-2-2x^2)>0 and we know that 2x-2-2x^2<0, \forall x \in \mathbb{R} due to the shape of the parabola and the fact that the discriminant is less than 0. So in order for our overall expression to be greater than zero, we must have (3x-2)(2x+1)<0 whereas you marked that it must be greater than 0 for some reason.

    Also in both methods you get the inequality (3x-2)(2x+1)<0 as you can see.
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    (Original post by FHL123)
    Drawing the curves doesn't help
    It doesn't help but I don't understand why the other meathod doesn't work
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    Could you take a look at this question please?
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    (Original post by FHL123)
    It doesn't help but I don't understand why the other meathod doesn't work
    (Original post by FHL123)
    Could you take a look at this question please?
    I have literally just pointed out where you went wrong...
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    (Original post by RDKGames)
    Actually both methods give the same answers. For method one, notice that you have  (3x-2)(2x+1)(2x-2-2x^2)>0 and we know that 2x-2-2x^2<0, \forall x \in \mathbb{R} due to the shape of the parabola and the fact that the discriminant is less than 0. So in order for our overall expression to be greater than zero, we must have (3x-2)(2x+1)<0 whereas you marked that it must be greater than 0 for some reason.

    Also in both methods you get the inequality (3x-2)(2x+1)<0 as you can see.
    Thank you- genius !!!
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    (Original post by RDKGames)
    I have literally just pointed out where you went wrong...
    Apologies- I saw that after I sent that post!
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    but if you say that the (2x-2-2x^2) is less than zero, for method two doesn't (3x-2)(2x+1) have to be greater than zero, so the overall expression remains negative, as a positive and a negative is a negative?
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    (Original post by FHL123)
    but if you say that the (2x-2-2x^2) is less than zero, for method two doesn't (3x-2)(2x+1) have to be greater than zero, so the overall expression remains negative, as a positive and a negative is a negative?
    For method two you multiplied both sides by -1 so the inequality sign changes and your quadratic becomes positive for all x as a result of distributing the -1 into it. So to have your overall expression to be negative, you need to multiply something positive (the quadratic) by something negative (the product of 2 linear factors)
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    (Original post by RDKGames)
    For method two you multiplied both sides by -1 so the inequality sign changes and your quadratic becomes positive for all x as a result of distributing the -1 into it. So to have your overall expression to be negative, you need to multiply something positive (the quadratic) by something negative (the product of 2 linear factors)
    For method two i just rearranged it so the whole expression was less than zero by taking the LHS away from the RHS- I didn'f multiply the expression by -1?
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    (Original post by FHL123)
    For method two i just rearranged it so the whole expression was less than zero by taking the LHS away from the RHS- I didn'f multiply the expression by -1?
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    (Original post by FHL123)
    For method two i just rearranged it so the whole expression was less than zero by taking the LHS away from the RHS- I didn'f multiply the expression by -1?
    Well that's what essentially happened.

    \displaystyle (3x-2)(2x+1)\underbrace{(2x-2-2x^2)}_{<0} > 0

    \displaystyle -(3x-2)(2x+1)\underbrace{(2x-2-2x^2)}_{<0} < 0

    \displaystyle (3x-2)(2x+1)\underbrace{(-2x+2+2x^2)}_{>0} < 0


    In essence, if I have x-y>0 then I can multiply both sides by -1 to get -x+y<0 which would be the same as moving x,y onto the RHS that gives you 0>y-x. Similar thing happening here with your approach.
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    (Original post by RDKGames)
    Well that's what essentially happened.

    \displaystyle (3x-2)(2x+1)\underbrace{(2x-2-2x^2)}_{<0} > 0

    \displaystyle -(3x-2)(2x+1)\underbrace{(2x-2-2x^2)}_{<0} < 0

    \displaystyle (3x-2)(2x+1)\underbrace{(-2x+2+2x^2)}_{>0} < 0


    In essence, if I have x-y>0 then I can multiply both sides by -1 to get -x+y<0 which would be the same as moving x,y onto the RHS that gives you 0>y-x. Similar thing happening here with your approach.
    Why is (2x^2 - 2x +2) greater than zero if the discriminant is less than zero?
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    (Original post by FHL123)
    Why is (2x^2 - 2x +2) greater than zero if the discriminant is less than zero?
    Discriminant being less 0 doesn't denote whether a quadratic is strictly positive or negative. It simply says that it doesn't cross the x-axis.
 
 
 
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