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Math Help - Roots of quadratic equation

  1. #1
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    Roots of quadratic equation

    The roots of the quadratic equation 16x^2 + 7x + 4 = 0 are \alpha^2 and \beta^2. Find two distinct quadratic equations whose roots are \alpha and \beta.

    My steps:

    Sum of roots, \alpha^2 + \beta^2 = \frac{-7}{16}
    Product of roots, \alpha^2\beta^2 = \frac{1}{4}

    Sum of new roots, \alpha + \beta = ... = \sqrt{\frac{-7}{16} + 2\sqrt\frac{1}{4}} = \pm\frac{3}{4}
    Product of new roots, \alpha\beta = ... = \sqrt{\frac{1}{4}} = \pm\frac{1}{2}

    But it turns out I'm partially correct, because while there is a \pm before \frac{3}{4}, there is not supposed to be a \pm before \frac{1}{2}.

    After staring at it for a couple minutes I'm guessing that since the original equation has a positive c/a, hence my new product of roots must be positive too. Am I right?
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  2. #2
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    Quote Originally Posted by fterh View Post
    The roots of the quadratic equation 16x^2 + 7x + 4 = 0 are \alpha^2 and \beta^2. Find two distinct quadratic equations whose roots are \alpha and \beta.

    My steps:

    Sum of roots, \alpha^2 + \beta^2 = \frac{-7}{16}
    Product of roots, \alpha^2\beta^2 = \frac{1}{4}

    Sum of new roots, \alpha + \beta = ... = \sqrt{\frac{-7}{16} + 2\sqrt\frac{1}{4}} = \pm\frac{3}{4}
    Product of new roots, \alpha\beta = ... = \sqrt{\frac{1}{4}} = \pm\frac{1}{2}

    But it turns out I'm partially correct, because while there is a \pm before \frac{3}{4}, there is not supposed to be a \pm before \frac{1}{2}.

    After staring at it for a couple minutes I'm guessing that since the original equation has a positive c/a, hence my new product of roots must be positive too. Am I right?
    Dear fterh,

    \alpha+\beta=\sqrt{(\alpha+\beta)^2}=\sqrt{\alpha^  2+\beta^2+2\alpha\beta}=\sqrt{-\frac{7}{16}+2\left(\frac{1}{2}\right)}=\frac{3}{4  }~not~\pm\frac{3}{4}
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  3. #3
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    Quote Originally Posted by Sudharaka View Post
    Dear fterh,

    \alpha+\beta=\sqrt{(\alpha+\beta)^2}=\sqrt{\alpha^  2+\beta^2+2\alpha\beta}=\sqrt{-\frac{7}{16}+2\left(\frac{1}{2}\right)}=\frac{3}{4  }~not~\pm\frac{3}{4}
    But the official answer in the book requires \pm\frac{3}{4} to arrive at, and it makes sense, because squaring both \frac{3}{4} and \frac{-3}{4} gets the same answer.
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  4. #4
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    Quote Originally Posted by fterh View Post
    But the official answer in the book requires \pm\frac{3}{4} to arrive at, and it makes sense, because squaring both \frac{3}{4} and \frac{-3}{4} gets the same answer.
    Dear fterh,

    When talking about the squareroot of a number we normally refer to the principle square root(the positive squareroot). Refer: Square root - Wikipedia, the free encyclopedia

    The question says to find two distinct quadratic equations. So, \alpha+\beta=\frac{3}{4}~and~\alpha\beta=\frac{1}{  2} gives one quadratic equation. Whereas \alpha+\beta=\frac{3}{4}~and~\alpha\beta=-\frac{1}{2} gives the other quadratic equation.

    Hope this helps.
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  5. #5
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    Quote Originally Posted by Sudharaka View Post
    Dear fterh,

    When talking about the squareroot of a number we normally refer to the principle square root(the positive squareroot). Refer: Square root - Wikipedia, the free encyclopedia

    The question says to find two distinct quadratic equations. So, \alpha+\beta=\frac{3}{4}~and~\alpha\beta=\frac{1}{  2} gives one quadratic equation. And \alpha\beta=-\frac{1}{2} gives the other quadratic equation.

    Hope this helps.
    I've discussed this extensively with a friend, and got it already But thanks for your help anyway. Although \alpha\beta cannot be \frac{-1}{2}, because if it were we'd have a problem finding the sum of new roots because we'd be trying to root a negative number then.
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