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mathmath333

  • one year ago

Solve for \(x\).

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} (x^2+3x+1)(x^2+3x-3)\geq 5\hspace{.33em}\\~\\ \end{align}}\)

  2. Loser66
    • one year ago
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    .

  3. ganeshie8
    • one year ago
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    One dumb method is to expand everything out and factor it again absorbing that 5 on right hand side

  4. ybarrap
    • one year ago
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    And that gives $$ (x-1) (x+1) (x+2) (x+4)-5 $$ Then find for what values of x is this greater or equal to 0

  5. ganeshie8
    • one year ago
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    Pretty sure you mean $$ (x-1) (x+1) (x+2) (x+4)\ge 0 $$ Then find for what values of x is this greater or equal to 0

  6. IrishBoy123
    • one year ago
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    .

  7. mathmath333
    • one year ago
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    how did u factor that \(\large \color{black}{\begin{align} (x-1) (x+1) (x+2) (x+4)\ge 0\hspace{.33em}\\~\\ \end{align}}\)

  8. dan815
    • one year ago
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    x(x+3) +1 x(x+3) -3|dw:1436812802011:dw|

  9. dan815
    • one year ago
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    |dw:1436812913492:dw|

  10. dan815
    • one year ago
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    u know the rate of increase of both parabolas so u see the first instance they pass pr0duct 5

  11. dan815
    • one year ago
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    then from axis of symmetry at 1.5 u can see the other end too

  12. anonymous
    • one year ago
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    @mathmath333 $$(x^2+3x+1)(x^2+3x-3)-5=(u+4)u-5=u^2+4u-5=(u+5)(u-1)$$for \(u=x^2+3x-3\)... and then we see that \((x^2+3x-3+5)(x^2+3x^2-3-1)=(x^2+3x+2)(x^2+3x-4)\) and both of these factor easily $$x^2+3x+2=(x+1)(x+2)\\x^2+3x-4=(x+4)(x-1)$$

  13. anonymous
    • one year ago
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    so for \(f(x)=(x+1)(x+2)(x+4)(x-1)\ge 0\) we know that the solution is going to be every other interval between the roots, so either $$(-\infty,-4]\cup[-2,-1]\cup[1,\infty)\\\text{or}\\ [-4,-2]\cup[-1,1]$$

  14. anonymous
    • one year ago
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    to determine which just test a point in one of the intervals: $$f(0)=(1)(2)(4)(-1)=-8<0$$so it follows the second intervals solve \(f(x)\le 0\) while the first solve \(f(x)\ge 0 \)

  15. mathmath333
    • one year ago
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    i d k why i am not able to see latex.

  16. dan815
    • one year ago
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    reload

  17. anonymous
    • one year ago
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  18. mathmath333
    • one year ago
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    reloaded twice.

  19. anonymous
    • one year ago
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    well, you don't need to test both intervals; you already know that since each root has multiplicity 1 the function is crossing between each so alternating between positive and negative; then you just test a point in any one of the intervals to distinguish which of these two solutions is for nonnegative \(f\) and which is for nonpositive \(f\)

  20. Loser66
    • one year ago
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    @oldrin.bataku Since I use table to determine it. It is like |dw:1436813611624:dw|

  21. Loser66
    • one year ago
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    |dw:1436813832253:dw|

  22. anonymous
    • one year ago
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    no I understand, I'm just saying that's more work than necessary, since you only need to check the sign of one of the intervals (not all of them) since we already determined the solution must either be \((-\infty,-4]\cup[-2,-1]\cup[1,\infty)\) or \([-4,-2]\cup[-1,1]\)

  23. anonymous
    • one year ago
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    checking the sign on say \([-2,-1]\) is enough to tell you the sign on \((-\infty,-4]\cup[-2,-1]\cup[1,\infty)\) since the function is alternating in sign between intervals so it must have the same sign for all of these, and the opposite sign for \([-4,-2]\cup[-1,1]\)

  24. Loser66
    • one year ago
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    Got you!!:)

  25. Loser66
    • one year ago
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    @oldrin.bataku Is there any other way to solve it?

  26. ybarrap
    • one year ago
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    The best way is the @oldrin.bataku approach, if you can see it. Otherwise, (the hard way) is to us the Rational Root Theorem, which gives you the possible roots. After multiplying everything out we get: $$ (x^2+3x+1)(x^2+3x-3)-5=x^4+6 x^3+7 x^2-6 x-8 $$ The possible roots are then, \(\pm1,\pm2,\pm4,\pm8\). Then using synthetic division, test each. https://en.wikipedia.org/wiki/Rational_root_theorem https://en.wikipedia.org/wiki/Synthetic_division I knew a most there was 1 positive root, using Descartes' rule of signs, that also helped - https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs I got lucky trying x=1,-1 first then everything else was easier because then I was just left with an easy quadratic to factor.

  27. freckles
    • one year ago
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    like the only other way I found is just a similar substitution but everything else is the same choose u=x^2+3x makes things factorable over the integers \[u=x^2+3x \\ (u+1)(u-3)-5 \ge 0 \\ (u^2-2u-3)-5 \ge 0 \\ u^2-2u-8 \ge 0 \\ (u-4)(u+2) \ge 0 \\ (x^2+3x-4)(x^2+3x+2) \ge 0 \\ (x-1)(x+4)(x+1)(x+2) \ge 0\]

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