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Math Help - quadractic inequality.

  1. #1
    Member integral's Avatar
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    quadractic inequality.

    My instructor told us, to solve a quadratic inequality factor out it's roots to set up a test interval on a number line. Then, test to see at what positions the inequality is true.

    My question:

    x^2+4x+4\geq 9<br />
    Descartes rule of signs:
    0 positive roots
    1 negative root


    rational root theorem:
    <br />
\textrm{roots}=\left [ -4,-2,-1  \right ]<br />
    Synthetic division:
    ............-4..0
    -4......1..4..4..
    ........1..0..4---- r=4\,\,\, \therefore not a root
    ............-2..-4
    -2.......1..4..4
    .........1..2..0-------- r=4\,\,\,\therefore a root!

    root=-2
    (x+2)(x+2)\geq 9

    But, this is a double root.
    And because it is a double root my interval is... not applicable besides \left ( -\infty, -2 \right ) or \left ( -2,\infty ) \right )

    But my textbook says the answer is:
    <br />
\left ( -\infty, -5 \right ]\cup \left [ 1, \infty \right )

    Any help on this problem?

    Also, if anyone has a more 'algebraic' way of solving a quadratic inequality, please post it. I hate guess and check problems.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by integral View Post
    My instructor told us, to solve a quadratic inequality factor out it's roots to set up a test interval on a number line. Then, test to see at what positions the inequality is true.

    My question:

    x^2+4x+4\geq 9<br />
    Descartes rule of signs:
    0 positive roots
    1 negative root


    rational root theorem:
    <br />
\textrm{roots}=\left [ -4,-2,-1 \right ]<br />
    Synthetic division:
    ............-4..0
    -4......1..4..4..
    ........1..0..4---- r=4\,\,\, \therefore not a root
    ............-2..-4
    -2.......1..4..4
    .........1..2..0-------- r=4\,\,\,\therefore a root!

    root=-2
    (x+2)(x+2)\geq 9

    But, this is a double root.
    And because it is a double root my interval is... not applicable besides


    But my textbook says the answer is:
    <br />
\left ( -\infty, -5 \right ]\cup \left [ 1, \infty \right )

    Any help on this problem?

    Also, if anyone has a more 'algebraic' way of solving a quadratic inequality, please post it. I hate guess and check problems.
    Hi integral,

    You first must solve the quadratic equation:

    x^2+4x+4=9

    x^2+4x-5=0

    (x+5)(x-1)=0

    r\:\:x=1" alt="x=-5 \:\r\:\:x=1" />

    Mark these two points on a number line. They will be part of your solution.

    Check a value in the interval to the left of -5 to see if it makes the original inequality true.

    You'll find that it does.

    Check a value between -5 and 1 to see if it makes the original inequality true..

    You'll find that it doesn't.

    Finally, check a value in the interval to the right of 1 to see if it makes the original inequality true.

    You'll find that it does.

    Conclusion:

    Everything from -5 to negative infinity works, including -5.

    Everything from 1 to positive infinity works, including 1.

    Mathematically:

    <br />
\left ( -\infty, -5 \right ]\cup \left [ 1, \infty \right )
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