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Math Help - Finding the roots of a Quadratic equation.

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    Finding the roots of a Quadratic equation.

    I am having a hard time with (4a-2x)^2= (5a-3x)(7x-5a). Should I use the Quadratic Formula?
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    Re: Finding the roots of a Quadratic equation.

    You should expand both sides of the equation, then collect like terms with one side being zero first.
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    Re: Finding the roots of a Quadratic equation.

    Quote Originally Posted by Berean View Post
    I am having a hard time with (4a-2x)^2= (5a-3x)(7x-5a). Should I use the Quadratic Formula?
    There is no common factor so, yes, go ahead and multiply out each side, put the quadratic in "standard form" and either factor (if you can), complete the square, or use the quadratic formula.
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    Re: Finding the roots of a Quadratic equation.

    Quote Originally Posted by MarkFL2 View Post
    You should expand both sides of the equation, then collect like terms with one side being zero first.
    Thankyou
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    Re: Finding the roots of a Quadratic equation.

    Thankyou
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    Re: Finding the roots of a Quadratic equation.

    Expanding both sides and the adding/subtracting until the equation = 0 (41a^2+25x^2-66ax = 0). I now have two numbers with an exponent of 2. Using the Quadratic equation example in my book ( ax^2 + bx +c = 0) only has one value with the exponent of two. Am I missing something?
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    MHF Contributor MarkFL's Avatar
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    Re: Finding the roots of a Quadratic equation.

    Assuming you are to solve for x, you then write the quadratic in x in standard form:

    25x^2-66ax+41a^2=0

    To avoid confusion between the a above and the a in the quadratic formula, one might substitute a = k.

    25x^2-66kx+41k^2=0

    Now identify:

    a=25

    b=-66k

    c=41k^2

    Now apply the quadratic formula, and remember to back-substitute for k when you're done.

    Edit: You may also observe that:

    25x^2-66ax+41a^2=0

    is factorable. 25 + 41 = 66...
    Last edited by MarkFL; December 30th 2011 at 11:36 AM.
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    Re: Finding the roots of a Quadratic equation.

    I have come up with an answer of .....a,(41/25)a. So the roots of (4a-2x)^2 = (5a-3x)(7x-5a) is a,(41/25)a?
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    MHF Contributor MarkFL's Avatar
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    Re: Finding the roots of a Quadratic equation.

    Yes. Did you notice you could factor the quadratic as:

    (x-a)(25x-41a)=0

    I didn't notice this actually until I saw the discriminant was a perfect square. That's one way to see if a quadratic has rational roots. Then I felt foolish for not noticing that 41 + 25 = 66.
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    Re: Finding the roots of a Quadratic equation.

    Quote Originally Posted by MarkFL2 View Post
    Yes. Did you notice you could factor the quadratic as:

    (x-a)(25x-41a)=0

    I didn't notice this actually until I saw the discriminant was a perfect square. That's one way to see if a quadratic has rational roots. Then I felt foolish for not noticing that 41 + 25 = 66.
    You will feel even more foolish when you realise 66 is not a square number although the discriminant is a square
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  11. #11
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    Re: Finding the roots of a Quadratic equation.

    Quote Originally Posted by e^(i*pi) View Post
    You will feel even more foolish when you realise 66 is not a square number although the discriminant is a square

    Won't you feel foolish when you realise you missed the point of 25 + 41 = 66

    ??
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