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Math Help - Discriminant

  1. #1
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    Discriminant

    Find the smallest value of the interger b for which 3x^2+bx+2 is positive for all values of x.

    My attempt

    since its always postive,

    b^2-4ac<0
    b^2-4(3)(2)<0
    b^2-24<0
    b^2<24
    b<\sqrt{24}


    however, the answer is -4
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  2. #2
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    Quote Originally Posted by Punch View Post
    Find the smallest value of the interger b for which 3x^2+bx+2 is positive for all values of x.

    My attempt

    since its always postive,

    b^2-4ac<0
    b^2-4(3)(2)<0
    b^2-24<0
    b^2<24
    b<\sqrt{24}


    however, the answer is -4


    Well...yes. you forgot that when taking square roots of real numbers we must deal with positive and/or negative values, so

    b^2<24\iff |b|<\sqrt{24}\iff -\sqrt{24}<b<\sqrt{24}\Longrightarrow the minimal integer (as required) value b can take is [-\sqrt{24}]+1=-4 ...

    Tonio
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  3. #3
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    Hello, Punch!

    Your work is almost correct.
    And be sure to answer the question.


    Find the smallest integer value of b for which 3x^2+bx+2 is positive for all values of x.

    My attempt: . b^2\:<\:24 \qquad{\color{blue}\Longrightarrow}\qquad {\color{blue}|b| \:<\:\sqrt{24}}

    However, the answer is -4

    We have: . -\sqrt{24} \:<\:b\:<\:\sqrt{24}

    . . . . . . . . -4.9 \:<\:b\:<\:4.9

    So b is between -4.9 and +4.9

    Code:
              ↓
      - - + o * * * * * * * * * * * * * * * * * o + - -
         -5  -4  -3  -2  -1   0   1   2   3   4   5

    So the least integer value of b is -4.



    Edit: ah, Tonio beat me to it . . . *sigh*
    .
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  4. #4
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    Thanks! you 2 are a great help heh!
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