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mathmath333

  • one year ago

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  1. mathmath333
    • one year ago
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    Find how many integers \(a\) are there such that \(\large \color{black}{\begin{align} 9x^2+3ax+a+5>0,\ x\in \mathbb{R} \hspace{.33em}\\~\\ \end{align}}\) for all values of \(x\). \(\large \color{black}{\begin{align} &a.) \ 7\hspace{.33em}\\~\\ &b.) \ 8\hspace{.33em}\\~\\ &c.) \ 9\hspace{.33em}\\~\\ &d.) \ 10\hspace{.33em}\\~\\ \end{align}}\)

  2. xapproachesinfinity
    • one year ago
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    i think you need to solve \[a^2-4a-20<0\] for that parabola to be >0 since the leading coefficient is >0 must be above x-axis so for that to be true the discriminant must be less than zero

  3. xapproachesinfinity
    • one year ago
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    so we need to draw the parabola a^2-4a-20 to see the range of integers where it is <0

  4. xapproachesinfinity
    • one year ago
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    according to this picture here https://www.desmos.com/calculator the range for a such that a^2-4a-20<0 -2.9<a<6.9 how many integers are there in that range

  5. xapproachesinfinity
    • one year ago
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    looks like 9 to me

  6. xapproachesinfinity
    • one year ago
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    took me to time to realize this!

  7. xapproachesinfinity
    • one year ago
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    but i might be wrong for my skills lol

  8. xapproachesinfinity
    • one year ago
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    @freckles

  9. mathmath333
    • one year ago
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    excellent!

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spraguer (Moderator)
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