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Thread: Discriminants and Real Solutions.

  1. #1
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    Discriminants and Real Solutions.

    Hello everyone, I'm in need of a bit of help.
    Could someone please explain the process of finding the discriminants and the number of real solutions to the quadratic equation of this expression:

    $\displaystyle 5x^2+7x+6=0$
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  2. #2
    Member SengNee's Avatar
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    Quote Originally Posted by tony_rose88 View Post
    Hello everyone, I'm in need of a bit of help.
    Could someone please explain the process of finding the discriminants and the number of real solutions to the quadratic equation of this expression:

    $\displaystyle 5x^2+7x+6=0$
    b^2-4ac
    =(7)^2-4(5)(6)
    =-71

    The discriminant is negative, therefore no real roots.
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  3. #3
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    For the polynomial $\displaystyle ax^{2} + bx + c = 0$, the discriminant is given as: $\displaystyle b^{2} - 4ac$. There are 3 cases in the value the discriminant can have and each tell you something about your quadratic/parabola:

    $\displaystyle b^{2} - 4ac > 0$ - The quadratic has two real roots (crosses x-axis twice)
    $\displaystyle b^{2} - 4ac = 0$ - The quadratic has exactly one real root (crosses x-axis once)
    $\displaystyle b^{2} - 4ac < 0$ - The quadratic has no real roots (never crosses x-axis)
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  4. #4
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    Thank you so much, both of you.
    I understand it now!

    But lets pretend that the Discriminants IS a positive number, how would I know how many Real Solutions there are?


    I have one more quick question totally unrelated while I have you here:

    I'm terrible at Simplifying Square Roots.
    Could you explain how to simplify this expression for me?

    $\displaystyle \sqrt(3y^2w^6) \sqrt(6yw^5)$ (not sure how to do the square root symbol on here)
    Assume that all variables represent positive real numbers.

    Thanks again!
    Last edited by tony_rose88; Mar 17th 2008 at 10:54 PM.
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  5. #5
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    Just two. For polynomials in the form of $\displaystyle ax^{2} + bx + c = 0$, you will not be able to draw the parabola crossing the x-axis more than twice.
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    \sqrt{xy} yields $\displaystyle \sqrt{xy}$.

    You can rewrite the square root as an exponent of one-half, i.e.:

    $\displaystyle \sqrt{a} = a^{\frac{1}{2}}$

    For your question, you can use your power rule to help simplify them after rewriting the square root as an exponent:

    $\displaystyle \left(a^{b}\right)^{c} = a^{b \cdot c}$


    Something that maybe of help ... In general: $\displaystyle a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}$
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  7. #7
    Senior Member topher0805's Avatar
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    $\displaystyle
    \sqrt{3y^2w^6} \sqrt {6yw^5}
    $

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}$

    Recall that:

    $\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

    So we have:

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}$

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}$
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  8. #8
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    Quote Originally Posted by topher0805 View Post
    $\displaystyle
    \sqrt{3y^2w^6} \sqrt {6yw^5}
    $

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}$

    Recall that:

    $\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

    So we have:

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}$

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}$
    Okay, I think I'm starting to get it, but could you explain to me where each of the 'parts' come from on this expression:
    $\displaystyle
    = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}$


    Also,
    Is there another way to write that without powers as fractions?

    I can't use fractions in my answer.
    All variables represent positive real numbers.

    Should I just turn them into decimals?

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y^{1.5}\cdot w^{5.5}$ ?
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  9. #9
    Senior Member topher0805's Avatar
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    No, don't put them into decimals.

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}
    $
    The $\displaystyle \sqrt{3}\cdot \sqrt{2}$ terms came from $\displaystyle \sqrt{6}$ .

    $\displaystyle \sqrt{6} = \sqrt{3\cdot 2} = \sqrt{3}\cdot \sqrt{2}$

    The last 4 terms came from this rule of exponents:

    $\displaystyle \sqrt{x} = x^{\frac {1}{2}}$

    Another of doing this without fractions as exponents is as follows:

    $\displaystyle
    = \sqrt{3}\cdot \sqrt{3}\cdot \sqrt{2}\cdot (y^2)^{\frac {1}{2}}\cdot y^{\frac {1}{2}}\cdot (w^6)^{\frac {1}{2}}\cdot (w^5)^{\frac {1}{2}}
    $

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot y\cdot \sqrt {y}\cdot w^{3}\cdot \sqrt {w^5}
    $
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  10. #10
    Senior Member topher0805's Avatar
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    Another way of writing the final step is:

    $\displaystyle
    3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}
    $

    $\displaystyle
    = 3\cdot \sqrt{2}\cdot \sqrt {y^3}\cdot \sqrt {w^{11}}
    $
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  11. #11
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    Ah, okay. I see!

    Thank you so much!
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