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Math Help - Finding Soloutions for Quadratics (Graphs)?

  1. #1
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    Finding Soloutions for Quadratics (Graphs)?

    Not sure how to do these two questions.

    The lines dont seem to exist.

    (e) x2 x 8 = 0

    (f) x2 + x = 4


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  2. #2
    No one in Particular VonNemo19's Avatar
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    Not enough info?

    All you have to do is solve for x in your quadratic equations and then substitute those values into the equation of your respaective line. This will give you the coordinates you seek. We call this in the biz, solving systems of equations.
    Last edited by VonNemo19; April 29th 2009 at 02:24 PM. Reason: messed up.
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    No one in Particular VonNemo19's Avatar
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    lsnguage

    By the way, write x squared like: x^2.

    it's less confusing.
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    Quote Originally Posted by VonNemo19 View Post
    All you have to do is solve for x in your quadratic equations and then substitute those values into the equation of your respaective line. This will give you the coordinates you seek. We call this in the biz, solving systems of equations.
    Hi. Could you explain this in more detail like give an example or something. We have to find the soloutions to the equations using the lines I.e where they intercept
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    No one in Particular VonNemo19's Avatar
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    Here ya Go

    O.K. Let's find the equation of the line that goes through the points

    (-2,0) and (0,2)

    By the point slope equation: y y1 = m(x x1),

    we see that (2-0)=m(0-(-2))
    2=2m
    m=1

    now the slope intercept form: y=mx+b
    using a point and the slope =1 we see that

    0=-2+b
    b=2 so the equation we seek is y=1x+2

    Now since y always equals y we solve the system

    Is that enough info?
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  6. #6
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    Not really. I have to find

    e) x2 x 8 = 0

    (f) x2 + x = 4



    How do I read off this graph for these two questions?

    Will probably just be easier to tell me the answers. Thanks


    Quote Originally Posted by VonNemo19 View Post
    O.K. Let's find the equation of the line that goes through the points

    (-2,0) and (0,2)

    By the point slope equation: y y1 = m(x x1),

    we see that (2-0)=m(0-(-2))
    2=2m
    m=1

    now the slope intercept form: y=mx+b
    using a point and the slope =1 we see that

    0=-2+b
    b=2 so the equation we seek is y=1x+2

    Now since y always equals y we solve the system

    Is that enough info?
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  7. #7
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    Talking

    Quote Originally Posted by Sailee316 View Post
    Will probably just be easier to tell me the answers.
    Yes, but how then would you learn?

    Try reviewing the solutions and suggestions you've received in various of your threads on this exercise, such as this and this.

    For instance, you have been told, for the first equation, to "Think about it! If y = 8 and y = x^2 - x, then x^2 - x = 8, etc." So think about what it means for two lines to be equal or to intersect. Then look at the graph and see how you can relate that to what is given to you in the picture.

    You might have to do a little thinking and manipulating, working with the concepts and the provided graphs, to arrive at your answer....

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  8. #8
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    Quote Originally Posted by Sailee316 View Post
    Not sure how to do these two questions.

    The lines dont seem to exist.

    (e) x2 x 8 = 0

    (f) x2 + x = 4


    What question are you talking about? Was the problem to solve those equations? I see that you have graphed y= x^2- x and y= 8. The roots of x^2- x- 8= 0 are the x-values where those graphs cross. It looks like one of the roots is between -3 and -2 and the other between 3 and 4. But what do y= x+ 2 and y= -2x+ 4 have to do with this problem?
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    Quote Originally Posted by HallsofIvy View Post
    What question are you talking about? Was the problem to solve those equations? I see that you have graphed y= x^2- x and y= 8. The roots of x^2- x- 8= 0 are the x-values where those graphs cross. It looks like one of the roots is between -3 and -2 and the other between 3 and 4. But what do y= x+ 2 and y= -2x+ 4 have to do with this problem?
    Hi. I did not draw this graph we were given it to answer about seven questions from. I answered the first 5 fine as I just had to look where they intercept. The other two (above) I think are much harder so I came on here hoping someone would give me the answers to these last two. Thanks!
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    Quote Originally Posted by Sailee316 View Post
    Hi. I did not draw this graph we were given it to answer about seven questions from. I answered the first 5 fine as I just had to look where they intercept. The other two (above) I think are much harder so I came on here hoping someone would give me the answers to these last two. Thanks!
    As you can see, there is some confusion as to what the questions actually are. You have not addressed the quite reaonsable questions asked by the previous poster, which makes it difficult to provide any further help at this stage.
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    Quote Originally Posted by mr fantastic View Post
    As you can see, there is some confusion as to what the questions actually are. You have not addressed the quite reaonsable questions asked by the previous poster, which makes it difficult to provide any further help at this stage.

    Use the graphs to find the solutions to:


    (e) x2 x 8 = 0

    (f) x2 + x = 4




    Graphs are at top of thread. I really need help with these questions. Thanks. We are suppose to find soloutions to these equations (co-ordinates) of where they intercept. x2 x refers to the graph y=x2 x. Thanks
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  12. #12
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    Quote Originally Posted by Sailee316 View Post
    Use the graphs to find the solutions to:


    (e) x2 – x – 8 = 0

    (f) x2 + x = 4




    Graphs are at top of thread. I really need help with these questions. Thanks. We are suppose to find soloutions to these equations (co-ordinates) of where they intercept. x2 – x refers to the graph y=x2 – x. Thanks
    Ahh! Thank you! x^2- x- 8= 0 is the same as x^2- x= 8 so solutions are the values of x where your graphs of y= x^2- x and y= 8 intersect. As I said before they appear to be a value between -3 and -2 and a value between 3 and 4.

    For x^2+ x= 4, to use the graphs you have, subtract 2x from both sides: x^2- x= 4- 2x= -2x+ 4. It looks like those graphs intersect for x between -3 and -2 and x between 1 and 2. If those graphs were on a computer or graphing calculator you could get more accurate values by enlarging the graphs or "zooming" in on the points.
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  13. #13
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    Quote Originally Posted by Sailee316 View Post
    Use the graphs to find the solutions to:


    (e) x2 – x – 8 = 0

    (f) x2 + x = 4




    Graphs are at top of thread. I really need help with these questions. Thanks. We are suppose to find soloutions to these equations (co-ordinates) of where they intercept. x2 – x refers to the graph y=x2 – x. Thanks
    x^2 - x - 8 = 0 \Rightarrow x^2 - x = 8.

    Geometrically, x^2 - x = 8 corresponds to solving y = x^2 - x and y = 8 simultaneously. The solution for x in this case is the same as the x-cordinate of the intersection point of the graphs of y = x^2 - x and y = 8.

    -------------------------------------------------------------------------------

    x^2 + x = 4 \Rightarrow x^2 - x + 2x = 4 \Rightarrow  x^2 - x = -2x + 4.

    Geometrically, x^2 - x = -2x + 4 corresponds to solving y = x^2 - x and y = -2x + 4 simultaneously. The solution for x in this case is the same as the x-cordinate of the intersection point of the graphs of y = x^2 - x and y = -2x + 4.

    -----------------------------------------------------------------------------------

    In both cases you can read off the x-coordinates of the intersection points using the graphs you've been given.
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    Hi. Thanks for your reply. I think
    I understand now. I can find the intersection points of the equations you have given me. Once I have found them do I have to change the x coordinate to a minus?? E.g 3 = -3, 4 = -4. How about y do I leave this positive or change to negative? Thanks
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  15. #15
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    Quote Originally Posted by Sailee316 View Post
    Hi. Thanks for your reply. I think
    I understand now. I can find the intersection points of the equations you have given me. Once I have found them do I have to change the x coordinate to a minus?? E.g 3 = -3, 4 = -4. How about y do I leave this positive or change to negative? Thanks
    The x-coordinate of the intersection points is all that you want. You don't change them to a minus and you don't care about the y-coordinates.
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