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Math Help - Solving Quadratic Equation With Equal Roots

  1. #1
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    Post Solving Quadratic Equation With Equal Roots

    Hi, I'm having a hard time dealing with this problem. It says "If (2m+1)(x^2+1)=(2m+5)x has equal roots, find the value of m"

    How do I solve this problem?
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  2. #2
    MHF Contributor red_dog's Avatar
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    (2m+1)x^2-(2m+5)x+2m+1=0

    If the roots are equal then the discriminant is 0.

    Find the discriminant and solve the quadratic in respect to m.
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  3. #3
    Member eXist's Avatar
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    Well to start I would distribute those and get:

    2mx^2 + 2m + x^2 + 1 = 2mx + 5x

     = (2m + 1)x^2 + (2m + 5)x + (2m + 1) = 0 (Set it equal to 0 since we need to find the roots.

    Apply the quadratic equation:

    \frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)}

    How ever, we need these roots to be equal to m, therefore we set this equation to m:

    \frac{-(2m + 5) \pm \sqrt{(2m + 5)^2 - 4(2m + 1)^2}}{2(2m + 1)} = m

    And from there, solve for m . Hope this helps, Chad.
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  4. #4
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    Quote Originally Posted by saberteeth View Post
    Hi, I'm having a hard time dealing with this problem. It says "If (2m+1)(x^2+1)=(2m+5)x has equal roots, find the value of m"

    How do I solve this problem?
    hi , let's change it into the general form : ax^2 + bx + c = 0 first

     (2m+1)(x^2+1) = (2m+5)x

     (2m+1)x^2 - (2m+5)x + (2m+1) =0

    Since the equation has two equal roots , the discriminant  b^2 - 4ac = 0

     [-(2m+5)]^2 - 4(2m+1)(2m+1) = 0

     (2m+5)^2 - 2^2(2m+1)^2 = 0

     [(2m+5)-(2)(2m+1)] [(2m+5) + (2)(2m+1) ] =0

     ( -2m + 3)( 6m + 7) = 0

    what's next ?
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  5. #5
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    thanx!
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