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Math Help - Solving Quadratic Equations Using The Square Root Principal

  1. #1
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    Solving Quadratic Equations Using The Square Root Principal

    I need help with this question: 5/2 = 6/(x+4)^2 (Rationalize the denominator) Thanks
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  2. #2
    Senior Member Stroodle's Avatar
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    The denominators already are rational.

    You can cross multiply then solve the equation. If that's what you want to do?

    \frac{5}{2}=\frac{6}{(x+4)^2}

    5(x+4)^2=12

    Can you take it from there?
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  3. #3
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    Quote Originally Posted by Stroodle View Post
    The denominators already are rational.

    You can cross multiply then solve the equation. If that's what you want to do?

    \frac{5}{2}=\frac{6}{(x+4)^2}

    5(x+4)^2=12

    Can you take it from there?
    Oh Wait I was FOILing 5(x+4)^2 instead of taking the square root, thanks. So after that i get 5x+20 = 2sqrt3 subtract 20 and divide by 5 to get x = (2sqrt3)/5 - 4 Is that right?
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  4. #4
    Senior Member Stroodle's Avatar
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    5(x+4)^2=12

    5(x+4)(x+4)=12

    5(x^2+8x+16)=12

    5x^2+40x+68=0

    Now you can either sub these values into the quadratic formula or complete the square to find the values of x.

    Do you know how to do this?
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  5. #5
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    I would use the quadratic formula but it says solve using the square root principal, is that just isolating x?
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  6. #6
    Senior Member Stroodle's Avatar
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    Oh, sorry I misunderstood. I'm not familiar with the square root principle. But I assume it's:

    5(x+4)^2=12

    (x+4)^2=\frac{12}{5}

    x+4=\pm\sqrt{\frac{12}{5}}

    x=\pm\sqrt{\frac{12}{5}}-4
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  7. #7
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    Yes that is it but you have to rationalize the denominator so would the final answer be x=2sqrt(15)/5-4?
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  8. #8
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    Quote Originally Posted by Stroodle View Post
    Oh, sorry I misunderstood. I'm not familiar with the square root principle. But I assume it's:

    5(x+4)^2=12

    (x+4)^2=\frac{12}{5}

    x+4=\pm\sqrt{\frac{12}{5}}

    x=\pm\sqrt{\frac{12}{5}}-4
    I'd make sure to rationalise the denominator...


    5(x + 4)^2 = 12

    (x + 4)^2 = \frac{12}{5}

    x + 4 = \pm\sqrt{\frac{12}{5}}

    x + 4 = \pm \frac{\sqrt{12}}{\sqrt{5}}

    x + 4 = \pm \frac{2\sqrt{3}}{\sqrt{5}}

    x + 4 = \pm \frac{2\sqrt{15}}{5}

    x = -4 \pm \frac{2\sqrt{15}}{5}

    x = \frac{-20 \pm \sqrt{15}}{5}.


    This should be the same answer you get if you use the Quadratic Formula...
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  9. #9
    Senior Member Stroodle's Avatar
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    Yep. The answer would be x=\pm\frac{2\sqrt{15}}{5}-4
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  10. #10
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    Oh yea plus or minus, thanks for your help I really appreciate it.
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  11. #11
    Senior Member Stroodle's Avatar
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    No probs. Finally got there in the end
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