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Math Help - Solving quadratics! help please

  1. #1
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    Smile Solving quadratics! help please

    Help would be greatly appreciated!

    1) Solve : 2/3(x^2 + 3x - 1)^2 = 3/8


    2) Solve the equation, by rearranging and completing the sqaure, finding the exact solutions - 6r^2 + 12r - pr^2 = 50

    Thank you in advance!!
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by steph_r View Post
    Help would be greatly appreciated!

    1) Solve : 2/3(x^2 + 3x - 1)^2 = 3/8
    \frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}

    (x^2 + 3x - 1)^2 = \frac{9}{16}

    x^2 + 3x - 1 = \pm \frac{3}{4}

    x^2 + 3x + \left ( -1 \pm \frac{3}{4} \right ) = 0

    So you have two equations to solve:
    x^2 + 3x - \frac{1}{4} = 0
    and
    x^2 + 3x - \frac{7}{4} = 0

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by steph_r View Post
    2) Solve the equation, by rearranging and completing the sqaure, finding the exact solutions - 6r^2 + 12r - pr^2 = 50
    -6r^2 + 12r - \pi r^2 = 50

    -(6 + \pi )r^2 - 12r = 50

    Get the coefficient of the leading term to be 1:
    r^2 + \frac{12}{6 + \pi } \cdot r = -\frac{50}{6 + \pi }

    Now complete the square:
    r^2 + \frac{12}{6 + \pi } \cdot r + \frac{144}{4(6 + \pi)^2} = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}

    \left ( r + \frac{12}{2(6 + \pi)} \right ) ^2 = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}

    I leave it to you to simplify and finish this out.

    -Dan
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  4. #4
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    Quote Originally Posted by steph_r View Post
    Help would be greatly appreciated!

    1) Solve : 2/3(x^2 + 3x - 1)^2 = 3/8


    2) Solve the equation, by rearranging and completing the sqaure, finding the exact solutions - 6r^2 + 12r - pr^2 = 50

    Thank you in advance!!

    1)
    \frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}
    (x^2 + 3x - 1)^2 = \frac{3}{8} \times \frac{3}{2}
    (x^2 + 3x - 1)^2 = \frac{9}{16}
    x^2 + 3x - 1 = \frac{3}{4}
    x^2 + 3x - 1 - \frac{3}{4} = 0
    x^2 + 3x -\frac{7}{4}=0
    (x - \frac{1}{2})(x + \frac{7}{2})
    \implies x = \frac{1}{2} \ \mathrm{or} \ x = \frac{-7}{2}

    Haven't done question 2 as it's been done when I clicked Preview Post. Question 1 was done too but since I had done it too, I thought I might submit the reply.
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  5. #5
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    Quote Originally Posted by Air View Post
    1)
    \frac{2}{3}(x^2 + 3x - 1)^2 = \frac{3}{8}
    (x^2 + 3x - 1)^2 = \frac{3}{8} \times \frac{3}{2}
    (x^2 + 3x - 1)^2 = \frac{9}{16}
    x^2 + 3x - 1 = \frac{3}{4}
    x^2 + 3x - 1 - \frac{3}{4} = 0
    x^2 + 3x -\frac{7}{4}=0
    (x - \frac{1}{2})(x + \frac{7}{2})
    \implies x = \frac{1}{2} \ \mathrm{or} \ x = \frac{-7}{2}

    Haven't done question 2 as it's been done when I clicked Preview Post. Question 1 was done too but since I had done it too, I thought I might submit the reply.
    Thank you very much!! I really appreciate it.
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  6. #6
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    Completing the square - extra help

    ...

    Quote Originally Posted by topsquark View Post
    -6r^2 + 12r - \pi r^2 = 50

    -(6 + \pi )r^2 - 12r = 50

    Get the coefficient of the leading term to be 1:
    r^2 + \frac{12}{6 + \pi } \cdot r = -\frac{50}{6 + \pi }

    Now complete the square:
    r^2 + \frac{12}{6 + \pi } \cdot r + \frac{144}{4(6 + \pi)^2} = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}

    Sorry, but may i ask how you went about the completing the square above? I dont quite understand

    \left ( r + \frac{12}{2(6 + \pi)} \right ) ^2 = -\frac{50}{6 + \pi } + \frac{144}{4(6 + \pi)^2}

    I leave it to you to simplify and finish this out.

    -Dan
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  7. #7
    Forum Admin topsquark's Avatar
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    The general idea for completing the square is this:

    You have something of the form
    x^2 + 2ax = c
    (You can always reduce your problem to this form.)

    Recall that x^2 + 2ax + a^2 = (x + a)^2

    So we need to find a, then add a^2 to both sides.

    In your case we have
    r^2 + \frac{12}{6 + \pi} \cdot r
    on the left hand side.

    So we need to solve
    2a = \frac{12}{6 + \pi} \implies a = \frac{12}{2(6 + \pi)}
    and then add
    a^2 = \frac{12^2}{2^2(6 + \pi)^2}
    to both sides.

    -Dan
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  8. #8
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    Thank you!

    ..

    Quote Originally Posted by topsquark View Post
    The general idea for completing the square is this:

    You have something of the form
    x^2 + 2ax = c
    (You can always reduce your problem to this form.)

    Recall that x^2 + 2ax + a^2 = (x + a)^2

    So we need to find a, then add a^2 to both sides.

    In your case we have
    r^2 + \frac{12}{6 + \pi} \cdot r
    on the left hand side.

    So we need to solve
    2a = \frac{12}{6 + \pi} \implies a = \frac{12}{2(6 + \pi)}
    and then add
    a^2 = \frac{12^2}{2^2(6 + \pi)^2}
    to both sides.

    -Dan

    Thank you! I understand now.. i really appreciate you help
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