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Math Help - modulus aka absolute value inequality

  1. #1
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    modulus aka absolute value inequality

    modulus aka absolute value inequality-untitled.gif my attempt i squared both sides led me to 4x^2+20x+25 is greater than x^2+4x+4
    3x^2+16x+21 is greater than 0 this lead to x is greater than -3 and less than -7/3 answer is x less than -3 and greater than -7/3
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    Re: modulus aka absolute value inequality

    Quote Originally Posted by abdulrehmanshah View Post
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ID:	30195 my attempt i squared both sides led me to 4x^2+20x+25 is greater than x^2+4x+4
    3x^2+16x+21 is greater than 0 this lead to x is greater than -3 and less than -7/3 answer is x less than -3 and greater than -7/3
    Have a look at this.
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    Re: modulus aka absolute value inequality

    i dont get i have seen this earlier but how to get to this answer what am i doing wrong
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    Re: modulus aka absolute value inequality

    Quote Originally Posted by abdulrehmanshah View Post
    i dont get i have seen this earlier but how to get to this answer what am i doing wrong
    $3x^2+16x+21>0\\(3x+7)(x+3)>0$

    Look at the sets $\left( { - \infty , - 3} \right),\;\left( { - 3,\frac{{ - 7}}{3}} \right),\;\left( {\frac{{ - 7}}{3},\infty } \right)$

    Which of those work.
    Thanks from abdulrehmanshah
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    Re: modulus aka absolute value inequality

    thanks bro i picture this as parabola
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    Re: modulus aka absolute value inequality

    I prefer to solve quadratic inequalities directly by completing the square, rather than looking at each possible solution set. (I know with higher polynomial and other inequalities this is the only possibility though)...

    $\displaystyle \begin{align*} 3x^2 + 16x + 21 &> 0 \\ x^2 + \frac{16}{3}x + 7 &> 0 \\ x^2 + \frac{16}{3}x + \left( \frac{8}{3} \right) ^2 - \left( \frac{8}{3} \right) ^2 + 7 &> 0 \\ \left( x + \frac{8}{3} \right) ^2 - \frac{64}{9} + \frac{63}{9} &> 0 \\ \left( x + \frac{8}{3} \right) ^2 - \frac{1}{9} &> 0 \\ \left( x + \frac{8}{3} \right) ^2 &> \frac{1}{9} \\ \sqrt{ \left( x + \frac{8}{3} \right) ^2 } &> \sqrt{ \frac{1}{9} } \\ \left| x + \frac{8}{3} \right| &> \frac{1}{3} \end{align*}$

    I'm sure you can go from here...
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