Results 1 to 11 of 11

Math Help - Discriminants and Real Solutions.

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    14

    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:

    5x^2+7x+6=0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member SengNee's Avatar
    Joined
    Jan 2008
    From
    Pangkor Island, Perak, Malaysia.
    Posts
    155
    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:

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

    The discriminant is negative, therefore no real roots.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    For the polynomial ax^{2} + bx + c = 0, the discriminant is given as: b^{2} - 4ac. There are 3 cases in the value the discriminant can have and each tell you something about your quadratic/parabola:

    b^{2} - 4ac > 0 - The quadratic has two real roots (crosses x-axis twice)
    b^{2} - 4ac = 0 - The quadratic has exactly one real root (crosses x-axis once)
    b^{2} - 4ac < 0 - The quadratic has no real roots (never crosses x-axis)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Mar 2008
    Posts
    14
    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?

    \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; March 17th 2008 at 11:54 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    Just two. For polynomials in the form of ax^{2} + bx + c = 0, you will not be able to draw the parabola crossing the x-axis more than twice.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    o_O
    o_O is offline
    Primero Espada
    o_O's Avatar
    Joined
    Mar 2008
    From
    Canada
    Posts
    1,407
    \sqrt{xy} yields \sqrt{xy}.

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

    \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:

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


    Something that maybe of help ... In general: a^{\frac{m}{n}} = \sqrt[n]{a^{m}} = \left(\sqrt[n]{a}\right)^{m}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    <br />
\sqrt{3y^2w^6} \sqrt {6yw^5}<br />

    <br />
= \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}

    Recall that:

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

    So we have:

    <br />
= \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}}

    <br />
= 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}

    <br />
= 3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Mar 2008
    Posts
    14
    Quote Originally Posted by topher0805 View Post
    <br />
\sqrt{3y^2w^6} \sqrt {6yw^5}<br />

    <br />
= \sqrt{3}\cdot \sqrt{y^2}\cdot \sqrt{w^6}\cdot \sqrt{6}\cdot \sqrt{y}\cdot \sqrt{w^5}

    Recall that:

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

    So we have:

    <br />
= \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}}

    <br />
= 3\cdot \sqrt{2}\cdot y^{\frac {2}{2}}\cdot y^{\frac {1}{2}}\cdot w^{\frac {6}{2}}\cdot w^{\frac {5}{2}}

    <br />
= 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:
    <br />
= \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?

    <br />
= 3\cdot \sqrt{2}\cdot y^{1.5}\cdot w^{5.5} ?
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    No, don't put them into decimals.

    <br />
= \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}}<br />
    The \sqrt{3}\cdot \sqrt{2} terms came from \sqrt{6} .

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

    The last 4 terms came from this rule of exponents:

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

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

    <br />
= \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}}<br />

    <br />
= 3\cdot \sqrt{2}\cdot y\cdot \sqrt {y}\cdot w^{3}\cdot \sqrt {w^5}<br />
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member topher0805's Avatar
    Joined
    Jan 2008
    From
    Vancouver
    Posts
    336
    Another way of writing the final step is:

    <br />
3\cdot \sqrt{2}\cdot y^{\frac {3}{2}}\cdot w^{\frac {11}{2}}<br />

    <br />
= 3\cdot \sqrt{2}\cdot \sqrt {y^3}\cdot \sqrt {w^{11}}<br />
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Mar 2008
    Posts
    14
    Ah, okay. I see!

    Thank you so much!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] No Real Solutions
    Posted in the Algebra Forum
    Replies: 8
    Last Post: November 8th 2011, 11:03 AM
  2. Replies: 1
    Last Post: August 11th 2009, 04:05 AM
  3. Real Solutions,
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 28th 2009, 07:18 AM
  4. real solutions question
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 28th 2008, 01:00 AM
  5. Real Solutions
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 26th 2006, 11:30 PM

Search Tags


/mathhelpforum @mathhelpforum