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    Can someone explain how the mark scheme seemingly skips over how the book says you are supposed to solve equations like this.

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    I thought that you would have to multiply both sides by  x^2 and  (x+3)^2 if you wanted to find the solutions for x where  x \in \mathbb{R} , such that you would not accidentally flip the inequality sign. Instead the mark scheme seems to just solve it the easy way.

    I got the same final solution though. I just had to solve a quantic instead of a quadratic (although it was not hard since the fact of x and (x+3) was already there).
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    (Original post by Louisb19)
    Can someone explain how the mark scheme seemingly skips over how the book says you are supposed to solve equations like this.

    Question
    Mark Scheme Answer

    I thought that you would have to multiply both sides by  x^2 and  (x+3)^2 if you wanted to find the solutions for x where  x \in \mathbb{R} , such that you would not accidentally flip the inequality sign. Instead the mark scheme seems to just solve it the easy way.

    I got the same final solution though. I just had to solve a quantic instead of a quadratic (although it was not hard since the fact of x and (x+3) was already there).
    I've only recently done this question! I'll be home in a bit and I'll take a picture and show you. Actually, I'll try explaining it. here's an easy way where you find all the critical values of the function (roots and vertical asymptotes)

    So, in this question, you identify all the critical values which is x=-2, x=6 (roots?) and V.A at x=-3, x=0.

    Then draw a table and split piecewise.

    So in region 1: x < -3: test a point in that region, let's say -4. If the inequality holds in that region, then a solution is x<-3. (I don't think it does here)

    Region 2: -3 < x < -2: test a point here, let's say -1.5, if the inequality holds (I think it does) then this region is a solution so -3 < x < -2 is a solution.

    Region 3: -2 < x < 0: test a point, etc...

    Region 4: 0 < x < 6: test a point etc...

    Region 5: x >6: test a point.

    Whilst this seems very cumbersome, if you draw up a table, it's done within half a minute.
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    (Original post by Zacken)
    I've only recently done this question! I'll be home in a bit and I'll take a picture and show you. Actually, I'll try explaining it. here's an easy way where you find all the critical values of the function (roots and vertical asymptotes)

    So, in this question, you identify all the critical values which is x=-2, x=6 (roots?) and V.A at x=-3, x=0.

    Then draw a table and split piecewise.

    So in region 1: x < -3: test a point in that region, let's say -4. If the inequality holds in that region, then a solution is x<-3. (I don't think it does here)

    Region 2: -3 < x < -2: test a point here, let's say -1.5, if the inequality holds (I think it does) then this region is a solution so -3 < x < -2 is a solution.

    Region 3: -2 < x < 0: test a point, etc...

    Region 4: 0 < x < 6: test a point etc...

    Region 5: x >6: test a point.

    Whilst this seems very cumbersome, if you draw up a table, it's done within half a minute.
    Yeah that makes sense, wasn't how it was taught in the book though. I guess you can do it either way.

    PS how do you remeber what the locus represented by

     arg(\dfrac{z - z_1}{z - z_2}) = \theta

    looks like. I'm trying to get an intuitive understanding of it so I can work it out by looking at it in the exam (I cannot be bothered to remember the long method of how to work out what it looks like with all those circle theorems). Danke schön Zain.
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    (Original post by Louisb19)
    Yeah that makes sense, wasn't how it was taught in the book though. I guess you can do it either way.
    Yep! Either way works. I like to do both + use a graphical calculator to check in the exam so I'm entirely sure of my answer.

    PS how do you remeber what the locus represented by

     arg(\dfrac{z - z_1}{z - z_2}) = \theta

    looks like. I'm trying to get an intuitive understanding of it so I can work it out by looking at it in the exam (I cannot be bothered to remember the long method of how to work out what it looks like with all those circle theorems). Danke schön Zain.
    I found the Examsolutions video fairly useful in understanding the concept of that locus.
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    (Original post by Louisb19)
    Yeah that makes sense, wasn't how it was taught in the book though. I guess you can do it either way.

    PS how do you remeber what the locus represented by

     arg(\dfrac{z - z_1}{z - z_2}) = \theta

    looks like. I'm trying to get an intuitive understanding of it so I can work it out by looking at it in the exam (I cannot be bothered to remember the long method of how to work out what it looks like with all those circle theorems). Danke schön Zain.
    Arg(z-a/z-b)=arg(z-a)-arg(z-b)
    So it's basically the locus of the points such that the angle made between the point and a and b is some specified angle - which is why you get an arc (major arc if angle is less than π/2 and a minor arc if angle is greater than π/2).
 
 
 
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