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anonymous

  • one year ago

Which quadratic equation is equivalent to (x + 2)2 + 5(x + 2) – 6 = 0? (u + 2)^2 + 5(u + 2) – 6 = 0 where u = (x – 2) u^2 + 4 + 5u – 6 = 0 where u = (x – 2) u^2 + 5u – 6 = 0 where u = (x + 2) u^2 + u – 6 = 0 where u = (x + 2)

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  1. misty1212
    • one year ago
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    lol HI again!

  2. anonymous
    • one year ago
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    hey girl!

  3. misty1212
    • one year ago
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    \[ (x + 2)^2 + 5(x + 2) – 6 = 0\] put \(\color{red}u=x+2\) get \[\huge\color{red}u^2+5\color{red}u-6=0\]

  4. anonymous
    • one year ago
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    or just go with C since it is always C

  5. misty1212
    • one year ago
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    lol ikr!

  6. anonymous
    • one year ago
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    When in doubt pick C lol.

  7. anonymous
    • one year ago
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    Which equation is quadratic in form? 4(x – 2)2 + 3x – 2 + 1 = 0 8x^5 + 4x^3 + 1 = 0 10x^8 + 7x^4 + 1 = 0 9x^16 + 6x^4 + 1 = 0

  8. anonymous
    • one year ago
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    @misty1212

  9. misty1212
    • one year ago
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    quadratic means hmm this is a trick

  10. misty1212
    • one year ago
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    i think they want you to say \(10x^8 + 7x^4 + 1 = 0\) is in "quadratic form" because you can make \(u=x^4\) and it becomes \[10u^2+7u+1=0\]

  11. misty1212
    • one year ago
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    the first one is in fact a quadratic equation, but C (as usual C) is in "quadratic form" go with that one

  12. anonymous
    • one year ago
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    What substitution should be used to rewrite 4x^12 – 5x6 – 14 = 0 as a quadratic equation? u = x^2 u = x^3 u = x^6 u = x^12

  13. anonymous
    • one year ago
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    @misty1212

  14. misty1212
    • one year ago
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    the middle term is has \(x^6\) so you can use \(u=x^6\)

  15. misty1212
    • one year ago
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    ooh what a surprise, C again...

  16. anonymous
    • one year ago
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    Thank you! What substitution should be used to rewrite 16(x^3 + 1)^2 – 22(x^3 + 1) – 3 = 0 as a quadratic equation? u = (x^3) u = (x^3 + 1) u = (x^3 + 1)^2 u = (x^3 + 1)^3

  17. misty1212
    • one year ago
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    wow not C

  18. misty1212
    • one year ago
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    you can use \(u=(x^3+1)\)

  19. anonymous
    • one year ago
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    What are the solutions of the equation x^4 + 3x^2 + 2 = 0? Use u substitution to solve.

  20. misty1212
    • one year ago
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    \[ x^4 + 3x^2 + 2 = 0\] put \[u=x^2\] get \[y=u^2+3u+2=0\] or \[(u+1)(u+2)=0\] but there will be no real solution only complex ones

  21. misty1212
    • one year ago
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    is there a typo in the question?

  22. anonymous
    • one year ago
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    A.) ±i (sqrt)2 and x = ±1 B) ±i (sqrt)2 and x = ±i C)±(sqrt)2 and x = ±i D)± (sqrt)2 and x = ±1

  23. misty1212
    • one year ago
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    oh they allow complex solutions ok

  24. anonymous
    • one year ago
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    yes it just took me a minute to type out! :]

  25. misty1212
    • one year ago
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    \[u+1=00\\ u=-1\\ ^2=-1\\ x=\pm i\] is one pair

  26. misty1212
    • one year ago
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    the other is \(x=\pm\sqrt{2}i\)

  27. misty1212
    • one year ago
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    looks like they put the \(i\) first , go with B

  28. anonymous
    • one year ago
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    What are the solutions of the equation x^4 – 5x^2 – 14 = 0? Use factoring to solve. Same as before.

  29. misty1212
    • one year ago
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    yeah same as before \[u=x^2\] get \[u^2-5u-14=0\] or \[(u-7)(u+2)=0\] so \[u=7\]or \[u=-2\]

  30. misty1212
    • one year ago
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    then \[x^2=7\\ x=\pm\sqrt7\] or \[x^2=-2\\ x=\pm\sqrt{2}i\]

  31. anonymous
    • one year ago
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    yayy thats what i had :)

  32. anonymous
    • one year ago
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    thank you!

  33. misty1212
    • one year ago
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    one to go?

  34. misty1212
    • one year ago
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    or was that it?

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