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Math Help - Quadratic Equations Using a Formula

  1. #1
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    Quadratic Equations Using a Formula

    Hi Happy New Year everyone hope you all have had a good one........ I'm doing Maths on New Years Day ..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....

    This is the equation :
    4-3x-2=0

    I use the formula :
    x = -b √ b - 4ac
    _____________
    2 a

    I substitute the letters for the numbers in the formula so that it looks like

    x = - - 3 √ -3 - 4 x 4 x -2
    _____________________
    2 x 4


    x = 3 + 4.80
    ________
    8

    or

    x = 3 - 4.80
    _________
    8

    Can someone please tell me where I'm going wrong

    I am stating roots correct to 2 dp
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  2. #2
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    Your a=4, b=-3 and c=-2. So, \sqrt{b^2 -4ac} = \sqrt{41} = 6.4. The answer has to be \frac{3 \pm \sqrt{41}}{8}.
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  3. #3
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    Quote Originally Posted by ExponentialP View Post
    Hi Happy New Year thank you
    everyone hope you all have had a good one........ I'm doing Maths on New Years Day ..... I'm having a bit of trouble with quadratic equations using a formula. I am using the formula to solve the equations but when I substitute the answers back in the equation the answers don't work....

    This is the equation :
    4-3x-2=0

    I use the formula :
    x = -b √ b - 4ac
    _____________
    2 a

    I substitute the letters for the numbers in the formula so that it looks like

    x = - - 3 √( -3) - 4 x 4 x -2
    _____________________
    2 x 4


    x = 3 + 4.80
    ________
    8

    or

    x = 3 - 4.80
    _________
    8

    Can someone please tell me where I'm going wrong

    I am stating roots correct to 2 dp
    See corrections.
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  4. #4
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    Yes thats what I done, but my answers still don't add up
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  5. #5
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    Quote Originally Posted by ExponentialP View Post
    Yes thats what I done, but my answers still don't add up
    If you calculate

    \sqrt{(-3)^2-4 \cdot 4 \cdot (-2)} = \sqrt{41}\approx 6.403

    but you still use \sqrt{23} \approx 4.8
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  6. #6
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    Thank u earbooth am I doing something wrong at this point




    x = 3 +6.40
    ________
    8

    or

    x = 3 - 6.40
    _________
    8
    Last edited by ExponentialP; January 1st 2010 at 10:22 AM.
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  7. #7
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    Quote Originally Posted by ExponentialP View Post
    Thank u earbooth am I doing something wrong at this point




    x = 3 +6.40
    ________
    8

    or

    x = 3 - 6.40
    _________
    8
    Both are equally valid solutions. Sub them back into the original equation if you want to be sure
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  8. #8
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    I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :

    2x - 7x = 3

    The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use

    a= 2 b= -7 c= 0 ????

    I then get

    x = - (-7) √ -7 - 4x2x0
    ____________________
    2 x 2

    The answer I get from this does not make sense, x = 7 √ -49
    ___________
    4
    please help and if possible could someone explain where I am going wrong in my calculation

    Thanks

    Happy New Year....... Again
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  9. #9
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    Quote Originally Posted by ExponentialP View Post
    I think I might be doing something wrong somewhere, I seem to be getting answers that don't make sense when I apply the formula to the equation, for instance take :

    2x - 7x = 3

    The exercise I am doing is asking me to solve this equation using a formula... I'm not sure if this is where I am going wrong but I use

    a= 2 b= -7 c= 0 ????

    I then get

    x = - (-7) √ -7 - 4x2x0
    ____________________
    2 x 2

    The answer I get from this does not make sense, x = 7 √ -49
    ___________
    4
    please help and if possible could someone explain where I am going wrong in my calculation

    Thanks

    Happy New Year....... Again
    If you want to apply the quadratic formula the equation must be in the form

    ax^2+bx+c=0

    Therefore you have to transform the equation first:

    2x^2 - 7x = 3~\implies~\boxed{2x^2-7x-3=0}

    So you have a = 2, b = -7, c = -3.
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  10. #10
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    Thank u !!!
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  11. #11
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    But I can't square root a negative number ??? What happens then ???
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  12. #12
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    Quote Originally Posted by ExponentialP View Post
    But I can't square root a negative number ??? What happens then ???
    2x^2 - 7x - 3 = 0

    b^2 - 4ac = (-7)^2 - 4(2)(-3) = 49+24 = 73

    x = \frac{7 \pm \sqrt{73}}{4}


    please review this site ...

    Discriminant in quadratic equations. This formula reveals critical information about the nature of the roots and solutions of a quadratic equation.
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