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anonymous

  • one year ago

Solve x2 − 7x + 12 = 0. x = −3, x = −4 x = 3, x = 4 x = 2, x = 6 x = −2, x = −6

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  1. anonymous
    • one year ago
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    factor the left part of the equation

  2. welshfella
    • one year ago
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    x^2 - 7x + 12 = 0 x^2 -4x - 3x + 12 = 0 now factor the above by grouping

  3. welshfella
    • one year ago
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    - that means factor x^2 - 4x and - 3x + 12 then simplify

  4. anonymous
    • one year ago
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    I agree

  5. anonymous
    • one year ago
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    @DarkMoonZ what did you get after following @welshfella directions

  6. anonymous
    • one year ago
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    Ok wait

  7. anonymous
    • one year ago
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    x (x - 4) and 3(4-x)

  8. anonymous
    • one year ago
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    Sorry I had been AFK lol and thanks @welshfella

  9. welshfella
    • one year ago
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    first one is write notice it -3 x + 12 negative 3

  10. anonymous
    • one year ago
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    oh ok

  11. anonymous
    • one year ago
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    so its -3(4-x)

  12. welshfella
    • one year ago
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    no its -3( x - 4) If you expand that you'll see it comes bakc to -3x + 12 so we got x(x - 4) - 3(x - 4) = 0

  13. anonymous
    • one year ago
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    Ok

  14. welshfella
    • one year ago
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    now e can simplify this further because there is (x - 4) in both parts

  15. welshfella
    • one year ago
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    so take x - 4 out and what have we got left?

  16. anonymous
    • one year ago
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    -3x=0

  17. anonymous
    • one year ago
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    or just X = -3

  18. welshfella
    • one year ago
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    no what i mean by taking x - 4 out is using it as a factor and what left becomes the other factor similar to 3 x + 3y - 3 is common to both 3x and 3y so we take 3 'out' and it becomes a factor 3x + 3y = 3 (x + y) The x+ y becomes the other factor.

  19. anonymous
    • one year ago
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    oh ok....

  20. welshfella
    • one year ago
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    x(x - 4) - 3(x - 4) = 0 (x - 4) is common so we take it out and whats left ( the x - 3 0 goes in the other parentheses so we have (x - 4)(x - 3) = 0

  21. welshfella
    • one year ago
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    now either x - 3 = 0 or x - 4 = 0 making x = 3 , 4

  22. anonymous
    • one year ago
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    Ok

  23. anonymous
    • one year ago
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    thanks

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