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anonymous

  • one year ago

Solve the quadratic equation y2 - 5y = -4 by factoring. {-2, 2} {2, 2} {-1, 4} {4, 1} i factor it, but i aint getting any of these?

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  1. anonymous
    • one year ago
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    @stacy.reed08

  2. anonymous
    • one year ago
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    @Nnesha

  3. Nnesha
    • one year ago
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    show ur work so i can find out the mistakes otherwise i have to start it ovr

  4. anonymous
    • one year ago
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    |dw:1439742404806:dw| thats all i got to,

  5. Nnesha
    • one year ago
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    cool now find two numbers when you multiply them you should get product of AC and when you add or subtract them you should get the middle term

  6. anonymous
    • one year ago
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    i did, i tried all the option and i couldnt find anything...

  7. anonymous
    • one year ago
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    i for sure thought it would be like (-1,4) but i didnt get 5

  8. Nnesha
    • one year ago
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    no i mean a-leading coefficient c=constant term |dw:1439742684150:dw| \[Ax^2+Bx+C=0\]

  9. Nnesha
    • one year ago
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    what two number you should multiply to get positive 4 but when you add them you should get -5

  10. anonymous
    • one year ago
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    -4 and -1

  11. Nnesha
    • one year ago
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    yes right so|dw:1439742854487:dw| set both parentheses equal to zero then solve for x

  12. Nnesha
    • one year ago
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    i mean for y*

  13. anonymous
    • one year ago
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    y=4 y=1

  14. Nnesha
    • one year ago
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    yep

  15. anonymous
    • one year ago
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    i was thinking it need to be (-4,-1) like just the numbers used to factor

  16. Nnesha
    • one year ago
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    no they are asking *solutions ) so you have to solve for the variable :)

  17. anonymous
    • one year ago
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    i would have like it to say *factor completely* bc when i was in public school, thats what it ment

  18. anonymous
    • one year ago
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    \[y^2-5y=-4\]\[y^2-5y+4=-4+4\]\[y^2-5y+4=0\]\[y^2-y-4y+4=0\]\[y(y-1)-4(y-1)=0\]\[(y-4)(y-1)=0\]and next the very last step...

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