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Math Help - Equations and Graph

  1. #1
    Member GAdams's Avatar
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    Equations and Graph

    I tried all sorts to get this question pasted or attached by itself but couldn't.

    I wonder if anyone would be kind enough go to the link and look at question number 22.

    It' page 20 -21 at this link: (If you have DSL it will take 1 min to download)

    http://www.aqa.org.uk/qual/gcse/qp-m...W-QP-NOV05.PDF

    Thanks.
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by GAdams View Post
    I tried all sorts to get this question pasted or attached by itself but couldn't.

    I wonder if anyone would be kind enough go to the link and look at question number 22.

    It' page 20 -21 at this link: (If you have DSL it will take 1 min to download)

    http://www.aqa.org.uk/qual/gcse/qp-m...W-QP-NOV05.PDF

    Thanks.
    You can "Print Screen" on that page, paste the image to MS Paint, save it to your computer, then attach the image to a post on these forums.
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  3. #3
    Member GAdams's Avatar
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    I tried printing and then scanning it and attaching it but as it happens, I don't have DSL (but I will soon).
    My connections is verry slow and it tries to attach the two pages but just sorta dies a slow whimpering death in the process.
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  4. #4
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by GAdams View Post
    I tried printing and then scanning it and attaching it but as it happens, I don't have DSL (but I will soon).
    My connections is verry slow and it tries to attach the two pages but just sorta dies a slow whimpering death in the process.
    On your keyboard, just above the "Insert" button, there's a button called "PrtScrn." Go to the page that you need help on in the PDF then click the "PrtScrn" button to "Print Screen" (this is basically a way of "copying" the screen's image which you can then "paste" into MS Paint).
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  5. #5
    Senior Member ecMathGeek's Avatar
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    I went ahead and loaded the PDF on my own. Just use the adivce I gave for future reference.
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  6. #6
    Senior Member ecMathGeek's Avatar
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    22 The grid on the opposite page shows graphs of a curve
     y = x^2 + 2x - 3

    and 3 straight lines

    y = x + 1
    y = -x - 2
    y = -x + 2

    You must use the graphs to answer the following questions.
    (a) Write down a pair of simultaneous linear equations that have a solution
    x = -1\frac{1}{2}, y=-\frac{1}{2}

    Look at the cooresponding graph. Go to the point (-1.5,-0.5) on the graph, and you will see that two of the lines interesect at this point. The two lines are y=x+1 and y=-x-2.

    (b) Write down and simplify a quadratic equation whose solutions are approximately -3.3 or 0.3. You must show clearly how to obtain your answer.

    For this, we want the x-intercepts of the quadratic equation to be (-3.3,0), (0.3,0). For this, we can set up a factored quadratic equation that has these solutions:

    y=(x+3.3)(x-0.3) \longleftarrow this equation has roots x=-3.3 and x=0.3 (which is what we want)
    y=x^2+3.3x-0.3x-3.3\cdot 0.3
    y=x^2+3x-0.99

    (I hope that makes sense to you. I'm not sure how far you are in that class.)

    (c) Write down the approximate solutions to the equation x^2 + x - 4 = 0. You must show clearly how you obtain your answer.

    I notice that this cannot be factored, so I'm thinking we should use the quadratic equation. However, I'm not sure if you've ever seen the quadratic equation. If you haven't, say so and I'm sure someone else can explain this problem differently.

    x=\frac{-1 \pm \sqrt{1^2-4(1)(-4)}}{2(1)}=\frac{-1 \pm \sqrt{17}}{2}

    x= 1.56 and x=-2.56
    Attached Thumbnails Attached Thumbnails Equations and Graph-p22.jpg  
    Last edited by ecMathGeek; May 27th 2007 at 10:57 AM.
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  7. #7
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    22 The grid on the opposite page shows graphs of a curve
     y = x^2 + 2x - 3

    and 3 straight lines

    y = x + 1
    y = -x - 2
    y = -x + 2

    You must use the graphs to answer the following questions.
    (a) Write down a pair of simultaneous linear equations that have a solution
    x = -1\frac{1}{2}, y=-\frac{1}{2}

    (b) Write down and simplify a quadratic equation whose solutions are approximately -3.3 or 0.3. You must show clearly how to obtain your answer.

    (c) Write down the approximate solutions to the equation x^2 + x - 4 = 0. You must show clearly how you obtain your answer.
    if there is anything you are unclear about, please ask.

    (a)
    y = x + 1 ..............(1)
    y = -x - 2 ...........(2)

    (b)

    x^2 + 2x - 3 = -x - 2

    \Rightarrow x^2 + 3x - 1 = 0

    By the quadratic formula:

    \Rightarrow x = \frac {-3 \pm \sqrt {9 + 4}}{2} = \frac {-3 \pm \sqrt {13}}{2} = 0.3 \mbox { or} -3.3


    (c)
    We could waste our time trying to figure this out from the graph (which is possible) or just solve the equation manually

    x^2 + x - 4 = 0

    \Rightarrow x = \frac {-1 \pm \sqrt {1 + 16}}{2} = \frac {-1 \pm \sqrt {17}}{2} = 1.56 \mbox { or} -2.56
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    You must use the graphs to answer the following questions.
    The first question is easily answered by using the graph, the other two are not. I'm not sure how you're supposed to use the graph for parts (b) and (c).
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  9. #9
    is up to his old tricks again! Jhevon's Avatar
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    Sorry ecMathGeek

    I saw you post the questions and not the answers, so I decided to post the answers. I should have known you would have come back to them--you can do such problems in your sleep I bet
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    The first question is easily answered by using the graph, the other two are not. I'm not sure how you're supposed to use the graph for parts (b) and (c).
    for (b) i used the graph by locating the solutions given as intersecting points for two of the graphs, then i equated them and simplified. i verified with the quadratic formula

    for (c), we could have constructed the given quadratic using two of the original graphs given. once we figure out which graphs to use, we could look at their intersecting points to find approximations to the solutions
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  11. #11
    Senior Member ecMathGeek's Avatar
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    Well, I can't figure out how to use a graph of equations that are not represented to find solutions to those equations.
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  12. #12
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Well, I can't figure out how to use a graph of equations that are not represented to find solutions to those equations.
    what do you mean?

    x^2 + x - 4 = x^2 + 2x - 3 - (x + 1)

    so look at the intersecting points for y = x^2 + 2x - 3 and y = x + 1, those will give our approx solutions
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  13. #13
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by Jhevon View Post
    for (b) i used the graph by locating the solutions given as intersecting points for two of the graphs, then i equated them and simplified. i verified with the quadratic formula

    for (c), we could have constructed the given quadratic using two of the original graphs given. once we figure out which graphs to use, we could look at their intersecting points to find approximations to the solutions
    Well done. I didn't even notice that.

    We're posting replies over each other. I saw this post after I posted the one you just responded to.
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  14. #14
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    Well done. I didn't even notice that.

    We're posting replies over each other. I saw this post after I posted the one you just responded to.
    yeah, i just noticed...you should slow down old lady (inside joke)
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  15. #15
    Member GAdams's Avatar
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    Quote Originally Posted by Jhevon View Post
    if there is anything you are unclear about, please ask.

    (a)
    y = x + 1 ..............(1)
    y = -x - 2 ...........(2)

    [/tex]
    I got the first equation by using the graph. I plotted -1 1/2 on the x axis and -1/2 on the y axis and drew the line. I then labelled it y=x+1.

    I can't work out how you got the second onethough.
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