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Solving quadratic equation

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    My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

    I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
    12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

    I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!
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    (Original post by miaofcourse)
    My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

    I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
    12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

    I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!
    Your expansion is wrong: (x+3)(3x-2) = x(3x-2) + 3(3x-2) = 3x^2 - 2x + 9x - 6 = 104

    Do you know that (a+b)(c+d) = ac + ad + bc + bd?
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    (Original post by miaofcourse)
    My question was; "A rectangle has sides of length x+3cm and 3x-2cm, given that the area of the rectangle is 104cm^2, find the value of x"

    I started out by doing the formula A=L x W and done x+3(3x-2) and by the end got
    12x^2-8x= 104 then I made in a quadratic by making it 12x^2-8x-104= .

    I am very stuck so I am pretty certain I have made a mistake, would any one be able to find my mistake and explain the process, thank you!!
    (x+3)(3x-2) = x(3x-2) + 3(3x-2) =\ ?
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    (Original post by Zacken)
    Your expansion is wrong: (x+3)(3x-2) = x(3x-2) + 3(3x-2) = 3x^2 - 2x + 9x - 6 = 104

    Do you know that (a+b)(c+d) = ac + ad + bc + bd?
    Thank you!! I see it now, in my working out I thought it was x+3x not x+3 haha
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    (Original post by miaofcourse)
    Thank you!! I see it now, in my working out I thought it was x+3x not x+3 haha
    Ahhh, glad it's sorted now!
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    (Original post by Zacken)
    Ahhh, glad it's sorted now!
    so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?
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    (Original post by miaofcourse)
    so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?
    When either bracket = 0 the whole thing = 0
    and we know x>0
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    (Original post by miaofcourse)
    so to simplify it would be 3x^2+7x-6 and then in brackets (3x+2)(x-3) how would that become the value of x?
    Hey! That's not correct.

    You started with (3x-2)(x+3) = 3x^2 + 7x - 6 = 104 then make it =0 to get 3x^2 + 7x -110 = 0 and now solve this by factorising or using the quadratic formula.
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    (Original post by Zacken)
    Hey! That's not correct.

    You started with (3x-2)(x+3) = 3x^2 + 7x - 6 = 104 then make it =0 to get 3x^2 + 7x -110 = 0 and now solve this by factorising or using the quadratic formula.
    So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!
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    (Original post by miaofcourse)
    So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!
    looks right
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    (Original post by miaofcourse)
    So it become (3x+22)(x-5) then x=-22 and x=5 and as after it says to work out the lenth and a length cannot be negative then x is 5?!
    Plug it back through the rectangle lengths to check.

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