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Math Help - An inequality

  1. #1
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    An inequality

    Hello!
    Solve the inequality 3x^2+4ix+5<0 , where i^2=-1.
    Thank You!
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  2. #2
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    Re: An inequality

    First, you have an inequality involving complex coefficients, but you are told it has to be less than a real number. Since complex numbers can NOT be ordered, you require putting in values which will give you a REAL solution.

    So that means that the expression \displaystyle \begin{align*} 3x^2 + 4i\,x + 5 \end{align*} needs to be real. So if we let \displaystyle \begin{align*} x = a + i\,b \end{align*} where \displaystyle \begin{align*} a,b \in \mathbf{R} \end{align*}, then that gives

    \displaystyle \begin{align*} 3x^2 + 4i\,x + 5 &= 3 \left( a + i\,b \right) ^2 + 4i \left( a + i\,b \right) + 5 \\ &= 3 \left[ a^2 - b^2 + i \left( 2\,a\,b \right) \right] + i \left( 4a \right) - 4b + 5 \\ &= 3a^2 - 3b^2 + i\left( 6\,a\,b \right) + i \left( 4a \right) - 4b + 5 \\ &= 3a^2 - 3b^2 - 4b + 5 + i \left( 4a + 6\,a\,b \right) \end{align*}

    For this expression to be real, that means

    \displaystyle \begin{align*} 4a + 6\,a\,b &= 0 \\ 2a \left( 2 + 3b \right) &= 0 \\ a = 0 \textrm{ or } b &= -\frac{2}{3}  \end{align*}


    Case 1: \displaystyle \begin{align*} a = 0 \end{align*} would give

    \displaystyle \begin{align*} 3a^2 - 3b^2 - 4b + 5 + i \left( 4a + 6\,a\,b \right) &= -3b^2 - 4b + 5 \end{align*}

    We require this expression to be \displaystyle \begin{align*} <0 \end{align*}, so we have

    \displaystyle \begin{align*} -3b^2 - 4b + 5 &< 0 \\ b^2 + \frac{4}{3}\,b - \frac{5}{3} &> 0 \\ b^2 + \frac{4}{3} \, b + \left( \frac{2}{3} \right) ^2 - \left( \frac{2}{3} \right) ^2 - \frac{5}{3} &> 0 \\ \left( b + \frac{2}{3} \right) ^2 - \frac{4}{9} - \frac{15}{9} &> 0 \\ \left( b + \frac{2}{3} \right) ^2 - \frac{19}{9} &> 0 \\ \left( b + \frac{2}{3} \right) ^2 &> \frac{19}{9} \\ \left| b + \frac{2}{3} \right| &> \frac{\sqrt{19}}{3} \\ b + \frac{2}{3} < -\frac{\sqrt{19}}{3} \textrm{ or } b + \frac{2}{3} &> \frac{\sqrt{19}}{3} \\ b < \frac{-2 - \sqrt{19}}{3} \textrm{ or } b &> \frac{-2 + \sqrt{19}}{3} \end{align*}


    Case 2: \displaystyle \begin{align*} b = -\frac{2}{3} \end{align*} would give

    \displaystyle \begin{align*} 3a^2 - 3b^2 - 4b + 5 + i \left( 4a + 6\,a\,b \right) &= 3a^2 - 3 \left( -\frac{2}{3} \right) ^2 - 4 \left( -\frac{2}{3} \right) + 5 \\ &= 3a^2 - 3 \left( \frac{4}{9} \right) + \frac{8}{3} + 5 \\ &= 3a^2 - \frac{4}{3} + \frac{8}{3} + \frac{15}{3} \\ &= 3a^2 + \frac{19}{3} \end{align*}

    We require this expression to be \displaystyle \begin{align*} < 0 \end{align*}, so we have \displaystyle \begin{align*} 3a^2 + \frac{19}{3} &< 0 \end{align*}, but since \displaystyle \begin{align*} a \in \mathbf{R} \end{align*} there are not any solutions.


    So that means that the solution to your equation \displaystyle \begin{align*} 3x^2 + 4i\,x + 5 < 0 \end{align*} is \displaystyle \begin{align*} x = 0 + i\,b \end{align*} where \displaystyle \begin{align*} b \in \left( -\infty , \frac{-2 - \sqrt{19}}{3} \right) \cup \left( \frac{ -2 + \sqrt{19}}{3} , \infty \right)  \end{align*}.
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  3. #3
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    Re: An inequality

    Dear Prove it,
    can we proceed like this: Let us put the equation equal to 0
    3x^2 + 4ix + 5 = 0 OR
    - 3 ( -ix ) ^ 2 + 4 ( ix ) + 5 = 0 Let ix = a We get
    3a^2 - 4a - 5 = 0, By using quadratic formula we get
    a = [ 4 =/- 4 sqrt ( 19 )] / 6 =[ 2 +/- 2 sqrt ( 19 )] / 3
    That means ix = [ 2 +/- 2 sqrt ( 19 )] / 3
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  4. #4
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    Re: An inequality

    Well that gives you where the function IS 0, but that's not what the question was asking. But it's a start
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