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Math Help - Having trouble with these two questions!

  1. #1
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    Having trouble with these two questions!

    "A container built to hold statures is based on a rectangular cuboid"


    The volume in cubic meters is given by the formula V=x^3 + 6x^2 + 8x

    a) Show that the sides have lengths x,x+2 and x+4 meters.
    b)An expresssion for the surface area of the container is 6x^2 + 24x +16
    If the surface are of the container is 142m^2, what is the volume?


    Last one:

    "A rectangular garden has a perimeter of 22m and an area of 21.84m^2. The width of the garden is x meters and the length is y meters.

    a) Write two equations for this situation in terms of x and y
    b) Solve the equations simultaneously and state the dimensions of the garden


    Any help would be great!
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  2. #2
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    Quote Originally Posted by krinkle View Post
    "A container built to hold statures is based on a rectangular cuboid"


    The volume in cubic meters is given by the formula V=x^3 + 6x^2 + 8x

    a) Show that the sides have lengths x,x+2 and x+4 meters.

    factor x^3 + 6x^2 + 8x

    b)An expresssion for the surface area of the container is 6x^2 + 24x +16
    If the surface are of the container is 142m^2, what is the volume?

    solve the quadratic equation 6x^2 + 24x + 16 = 142 for x ... then substitute the solution that makes sense in the context of the problem into the expression for V.


    Last one:

    "A rectangular garden has a perimeter of 22m and an area of 21.84m^2. The width of the garden is x meters and the length is y meters.

    a) Write two equations for this situation in terms of x and y

    2(x+y) = 22

    xy = 21.84


    b) Solve the equations simultaneously and state the dimensions of the garden

    do it.
    ...
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  3. #3
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    2(x+y) = 22

    xy = 21.84

    What are the answers to these?


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  4. #4
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    Also how to solve quadratic equation 6x^2 + 24x + 16 = 142 for x ... I thought the formula always ended in 0?
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  5. #5
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    Quote Originally Posted by krinkle View Post
    Also how to solve quadratic equation 6x^2 + 24x + 16 = 142 for x ... I thought the formula always ended in 0?
    try subtracting 142 from both sides and then using the quadratic formula
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    Quote Originally Posted by krinkle View Post
    2(x+y) = 22
    x+y =11
    y=11-x

    xy = 21.84
    x(11-x)=21.84

    Can you continue from here?




    .
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  7. #7
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    Quote Originally Posted by linalg123 View Post
    .
    Lol,
    No... does y = 11-x too?
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  8. #8
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    Quote Originally Posted by linalg123 View Post
    try subtracting 142 from both sides and then using the quadratic formula
    Both sides of what?
    Deeply confused!
    any solutions so i can work backwards?
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  9. #9
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    Quote Originally Posted by krinkle View Post
    Also how to solve quadratic equation 6x^2 + 24x + 16 = 142 for x ... I thought the formula always ended in 0?
    subtracting 142 from both sides of the equation gives us
    6x^2 + 24x + 16 - 142 = 142 - 142
    6x^2 + 24x - 126 = 0
    taking a common factor of 6 gives us 6( x^2 + 4x - 21) = 0
    this is now in the correct form to use the quadratic equation. Can you do this?

    2(x+y) = 22
    x+y =11
    y=11-x

    xy = 21.84
    x(11-x)=21.84
    11x - x^2 = 21.84
    x^2 - 11x + 21.84 = 0
    this is also now in the correct form to use the quadratic equation
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  10. #10
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    Quote Originally Posted by linalg123 View Post
    subtracting 142 from both sides of the equation gives us
    6x^2 + 24x + 16 - 142 = 142 - 142
    6x^2 + 24x - 126 = 0
    taking a common factor of 6 gives us 6( x^2 + 4x - 21) = 0
    this is now in the correct form to use the quadratic equation. Can you do this?

    2(x+y) = 22
    x+y =11
    y=11-x

    xy = 21.84
    x(11-x)=21.84
    11x - x^2 = 21.84
    x^2 - 11x + 21.84 = 0
    this is also now in the correct form to use the quadratic equation

    For the first quadratic im getting -7 and 3 - which one is right?
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  11. #11
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    Quote Originally Posted by krinkle View Post
    For the first quadratic im getting -7 and 3 - which one is right?
    remember x is a length. It doesn't make sense to have a negative length
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  12. #12
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    2(x+y) = 22
    x+y =11
    y=11-x



    however, what are the values for x and y?


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  13. #13
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    Quote Originally Posted by krinkle View Post
    2(x+y) = 22
    x+y =11
    y=11-x



    however, what are the values for x and y?


    subbing this into your other equation xy= 21.84 gives
    xy = 21.84
    x(11-x)=21.84
    11x - x^2 = 21.84
    x^2 - 11x + 21.84 = 0 (which has the form a(x^2) + bx + c = 0)

    so you use the quadratic formula to solve for x. Do you know how to do this?
    once you have solved for x, you plug that value into x+y=11 and solve for y.
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