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anonymous

  • one year ago

Help me, please?

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  1. anonymous
    • one year ago
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    A quadratic equation is shown below: 3x2 − 16x + 2 = 0 Part A: Describe the solution(s) to the equation by just determining the radicand. Show your work. (5 points) Part B: Solve 9x2 + 3x − 2 = 0 by using an appropriate method. Show the steps of your work, and explain why you chose the method used. (5 points)

  2. anonymous
    • one year ago
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    I need help finding the radicand.

  3. anonymous
    • one year ago
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    @peachpi Do you think you could help me with this one?

  4. anonymous
    • one year ago
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    i am sure we can do this

  5. anonymous
    • one year ago
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    Awesome, thank you!

  6. anonymous
    • one year ago
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    \[3x^2 − 16x + 2 = 0 \] is the equation right?

  7. anonymous
    • one year ago
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    'Tis correct~

  8. anonymous
    • one year ago
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    ok the radicand of \[ax^2+bx+c=0\] is \(b^2-4ac\)

  9. anonymous
    • one year ago
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    but that does not give you the actual solution the actual solutions are \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]so lets compute it

  10. anonymous
    • one year ago
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    actually i see is does not say "find the solutions" it says "Describe the solution(s)" the radicand in this case is \[(-6)^2-4\times 3\times 2=232\]

  11. anonymous
    • one year ago
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    since that is a positive number, but not a perfect square (like say 25 or 16) that means there are two real but irrational solutions

  12. anonymous
    • one year ago
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    Okay, so we'll have to find those irrational solutions, right? But how?

  13. anonymous
    • one year ago
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    they do not ask you to find them, but we can

  14. anonymous
    • one year ago
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    solution is \[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] lets plug in the numbers and see what we get

  15. anonymous
    • one year ago
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    we already know \(b^2-4ac=232\)

  16. anonymous
    • one year ago
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    plugging them in give \[\frac{16\pm\sqrt{232}}{2\times 3}\]

  17. anonymous
    • one year ago
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    then some more arithmetic is needed

  18. anonymous
    • one year ago
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    Okay, so for the top part does that mean you could add or subtract or is there a specific one you're supposed to use?

  19. anonymous
    • one year ago
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    it is a short hand for saying there are two different solution one with the plus, and another with the minus

  20. anonymous
    • one year ago
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    Oh alright, that makes sense. But how would you solve the top part? We've already established that the 232 ends up two real irrational numbers, if we're trying to figure those out now, what do we do on the top?

  21. anonymous
    • one year ago
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    we can to that but i want to stress that the question does not ask for the actual solutions, only to "describe " them the description is two real irrational solutions now lets find them

  22. anonymous
    • one year ago
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    Right, okay.

  23. anonymous
    • one year ago
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    first of \[232=4\times 58\] so \[\sqrt{232}=\sqrt{4\times 58}=\sqrt{4}\sqrt{58}=2\sqrt{58}\]\]

  24. anonymous
    • one year ago
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    therefore \[\frac{16\pm\sqrt{232}}{2\times 3}=\frac{16\pm2\sqrt{58}}{6}\]

  25. anonymous
    • one year ago
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    now cancel a 2 top and bottom to finish with \[\frac{8\pm\sqrt{58}}{3}\]

  26. anonymous
    • one year ago
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    since 58 is not a prefect square, \(\sqrt{58}\) is irrational

  27. anonymous
    • one year ago
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    Oh okay, I see that now.

  28. anonymous
    • one year ago
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    that is your final answer to the first one second one is easier

  29. anonymous
    • one year ago
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    Hopefully, because I'm terrible at those cx

  30. anonymous
    • one year ago
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    the second one factors

  31. anonymous
    • one year ago
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    Oh so its just a factoring problem?

  32. anonymous
    • one year ago
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    yeah pretty much

  33. anonymous
    • one year ago
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    \[ 9x^2 + 3x − 2 = 0 \\ (3x+1)(3x-2)=0\] set each factor equal to zero and solve for \(x\)

  34. anonymous
    • one year ago
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    Is that the finished product?

  35. anonymous
    • one year ago
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    no that is just what you get when you factor to finish solving for \(x\) solve\[3x+1=0\] and also \[3x-2=0\]

  36. anonymous
    • one year ago
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    The first one would be \[-\frac{ 1 }{ 3 }\] Right?

  37. anonymous
    • one year ago
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    And the second one \[\frac{ 2 }{ 3 }\]

  38. anonymous
    • one year ago
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    yup

  39. anonymous
    • one year ago
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    oh wait a second

  40. anonymous
    • one year ago
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    dang i made a mistake sorry

  41. anonymous
    • one year ago
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    Its alright ^.^

  42. anonymous
    • one year ago
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    it is not \[(3x+1)(3x-2)=0\] it is \[(3x-1)(3x+2)=0\]

  43. anonymous
    • one year ago
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    so solutions are \(\frac{1}{3}\) and \(-\frac{2}{3}\)

  44. anonymous
    • one year ago
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    Alright cool, so that's it?

  45. anonymous
    • one year ago
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    yup thats it

  46. anonymous
    • one year ago
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    Thank you so much! That helped so much!

  47. anonymous
    • one year ago
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    yw

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