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anonymous

  • one year ago

FACTOR 2x^2 + x - 28 = 0 There should be 2 x's. I am stuck at 2(x^2+x-14)=0

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  1. anonymous
    • one year ago
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    @haleyelizabeth2017 Lists steps

  2. haleyelizabeth2017
    • one year ago
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    Wait, it was originally \(2x^2+2x-28?\)

  3. haleyelizabeth2017
    • one year ago
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    What you have is the simplest it can go.

  4. anonymous
    • one year ago
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    Originally 2x^2+x-28=0

  5. haleyelizabeth2017
    • one year ago
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    Okay, then what you have so far is not correct.

  6. anonymous
    • one year ago
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    oh yea i made a mistake, i thought the GCF was 2 when it can't be

  7. anonymous
    • one year ago
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    what can be mult. to make -56 but added to make 1

  8. haleyelizabeth2017
    • one year ago
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    -7 and 4 multiply to get -28, okay? So then...if one side is (2x+/-_)(x+/-_) we can plug in those two numbers to try and see which works.

  9. anonymous
    • one year ago
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    im guessing that another way to do it Eli?

  10. haleyelizabeth2017
    • one year ago
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    So...let's try (2x-7)(x+4) first. We get \(2x^2+8x-7x-28\) which simplified, would work. So, (2x-7)(x+4) is the correct.

  11. haleyelizabeth2017
    • one year ago
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    correct answer* ^^

  12. anonymous
    • one year ago
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    so 2x-7=0 is x=7/2 or 3 1/2

  13. anonymous
    • one year ago
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    and x=-4

  14. haleyelizabeth2017
    • one year ago
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    Normally, it is left as an improper fraction, so x=7/2 is better. and x=-4 is correct as well.

  15. haleyelizabeth2017
    • one year ago
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    You could have also used the quadratic formula.

  16. haleyelizabeth2017
    • one year ago
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    It is a lot easier. Or, even complete the square works too

  17. haleyelizabeth2017
    • one year ago
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    \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] \[ax^2+bx+c=0\]

  18. anonymous
    • one year ago
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    Ok, i got a question... the formula is ax^2+bx+c... i was told that you mult. a*c and the number that you get has to be mult. by x number and x number, but those two number must also be able to be added together to equal B... did that make sense

  19. anonymous
    • one year ago
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    Is there an exception to this??

  20. haleyelizabeth2017
    • one year ago
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    Huh?

  21. haleyelizabeth2017
    • one year ago
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    You don't have to do anything to the a b or c. just plug the coefficients into the quadratic formula and solve. Lets use your problem for example. We were given \[2x^2+x-28\] A would be 2, b would be 1, and c would be -28.

  22. haleyelizabeth2017
    • one year ago
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    \[\frac{ -(1) \pm \sqrt{(1)^2-4(2)(-28)} }{ 2(2) }\]\[\frac{ -1 \pm \sqrt{1+224} }{ 4 }\]and so on

  23. haleyelizabeth2017
    • one year ago
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    Did that make sense?

  24. Hero
    • one year ago
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    @Beleaguer did you figure out What the two numbers are?

  25. Hero
    • one year ago
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    In other words, can you solve m+n =1 mn =-56

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