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I have a test tomorrow and can't solve this problem

  1. Jul 23, 2006 #1
    Can anyone please help? :cry:

    2x² - x + 1 < 9


    Edit: This is what I've done so far

    2x² - x - 8 < 0
    2 [ x² - 1/2x - 4] < 0
    2(x - 1/2)² - 17/4 < 0
     
    Last edited: Jul 23, 2006
  2. jcsd
  3. Jul 23, 2006 #2

    quasar987

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    This line is incorrect:

    2 [ x² - 1/2x - 4] = 2(x - 1/2)² - 17/4

    Start by getting the right one, then we'll go foward.
     
  4. Jul 23, 2006 #3
    Then should it be: 2 [ x² - 1/2x - 4] = 2(x - 1/2)² + 15/4

    :confused: I'm really lost :frown:
     
  5. Jul 23, 2006 #4

    quasar987

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    The formula you should memorize if you have difficulty with this is

    [tex]ax^2+bx+c= a\left[ \left( x+\frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} \right] [/tex]

    Apply this formula to your equation. (i.e. plug a=2, b=-1 and c=-8)
     
  6. Jul 23, 2006 #5
    another method is to solve [itex]2x^{2} - x - 8 = 0[/itex] by dividing both sides by 2 and solve the quadratic equation using

    [tex]x = \frac {-b ^{+}_{-} {\sqrt {b^{2} - 4ac}}} {2a}[/tex]

    and then investigate for what x this inequality is valid.
     
    Last edited: Jul 23, 2006
  7. Jul 24, 2006 #6

    VietDao29

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    Uhmm, let f(x) = ax2 + bx + c, and let the equation f(x) = 0 has 2 real roots x1, and x2, such that x1 < x2. We choose a arbitrary x0 value. We have:
    [tex]af(x_0) < 0 \Leftrightarrow x_0 \in ]x_1 , \ x_2[[/tex]
    [tex]af(x_0) > 0 \Leftrightarrow x_0 \notin ]x_1 , \ x_2[[/tex]
    Or in other words:
    x0 is in the interval ]x1, x2[ if and only if the leading coefficient a, and f(x0) takes different signs, i.e if a is positive, then f(x0) is negative, and vice versa.
    x0 is not in the interval ]x1, x2[ if and only if the leading coefficient a, and f(x0) takes the same sign, i.e if a is positive, then f(x0) is also positive, and vice versa.
    ---------------
    I'll give you an example:
    Example:
    Solve the inequality:
    x2 + 5x + 4 < 0
    <=> (x + 4) (x + 1) < 0
    -4, and -1 are the 2 roots for this equation.
    Let f(x) = x2 + 5x + 4
    The leading coefficient of f(x) is 1 > 0
    And you have to solve for f(x) < 0, i.e, you will have to find all the value x such that f(x) and the coefficient take different signs, so the solution is
    <=> -4 < x < -1
    Can you get this? :)
     
    Last edited: Jul 24, 2006
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