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Sunshine447 Group Title

Solve the equation by factoring: x2 - 7x + 12 = 0 A) x = 3 or 4 B) x = -3 or 4 C) x = 3 or -4 D) x = -3 or -4

  • one year ago
  • one year ago

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  1. Sunshine447 Group Title
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    please don't just give me the answer.... I want to actually try to learn how to do it. Also, know how to do this the "plug and play" method, but I need to know how to do it when it's not multiple choice. Thanks!

    • one year ago
  2. jiteshmeghwal9 Group Title
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    sorry dude

    • one year ago
  3. Sunshine447 Group Title
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    why?

    • one year ago
  4. jiteshmeghwal9 Group Title
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    first factorize the L.H.S. & then gt ur answer

    • one year ago
  5. Sunshine447 Group Title
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    LHS?

    • one year ago
  6. ParthKohli Group Title
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    Hi, Sunshine! Would you like to know the whole process from the start?

    • one year ago
  7. Sunshine447 Group Title
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    yes please! Thanks Spiderman! haha

    • one year ago
  8. Yahoo! Group Title
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    Sum = -7 Product = 12 two numbers -4 and -3 x^2 -4x - 3x + 12 =0 x(x-4) -3(x-4) (x-4) (x-3) x = 3 or x = 4

    • one year ago
  9. ParthKohli Group Title
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    All right then. Do you know what the standard form of a quadratic equation is?

    • one year ago
  10. Yahoo! Group Title
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    @Sunshine447 Did u understand!

    • one year ago
  11. Sunshine447 Group Title
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    @ParthKohli yes

    • one year ago
  12. ParthKohli Group Title
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    Okay, what is it?

    • one year ago
  13. Sunshine447 Group Title
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    ax + bx + c = 0 right?

    • one year ago
  14. ParthKohli Group Title
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    \(ax^2 + bx + c = 0\)* What you first do is find the product of \(a\) and \(c\). What are \(a\) and \(c\) here?

    • one year ago
  15. Yahoo! Group Title
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    @Sunshine447 Did u understand my method????

    • one year ago
  16. Sunshine447 Group Title
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    a=1 and c=12

    • one year ago
  17. ParthKohli Group Title
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    Right! What's the product of 1, 12?

    • one year ago
  18. Sunshine447 Group Title
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    12

    • one year ago
  19. ParthKohli Group Title
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    Correct! :) Now you have to find two numbers that add to get \(b\) and multiply to get the product of \(a\) and \(c\).

    • one year ago
  20. ParthKohli Group Title
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    Do you know two numbers such that: Their sum is -7. Their product is 12. ?

    • one year ago
  21. Sunshine447 Group Title
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    no

    • one year ago
  22. ParthKohli Group Title
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    Hmm, you may find them by trial and error. Note that a negative number times a negative number is a positive number, so those numbers must be negative if the sum of negative.

    • one year ago
  23. Sunshine447 Group Title
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    -3 and -4

    • one year ago
  24. ParthKohli Group Title
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    You got it! Now, we will 'split' the middle term with these coefficients. \(-7x = -3x - 4x \Longrightarrow x^2 - 7x + 12 = x^2 - 3x - 4x + 12\)

    • one year ago
  25. ParthKohli Group Title
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    Do you get it till this point?

    • one year ago
  26. Sunshine447 Group Title
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    yeah i think so

    • one year ago
  27. ParthKohli Group Title
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    Okay, now we basically factor by 'grouping'. What you must do is factor the first two and last two terms. \(x^2 - 3x = x(x - 3)\) \(-4x + 12 = -4(x - 3)\) Get it till here?

    • one year ago
  28. Sunshine447 Group Title
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    yeah

    • one year ago
  29. ParthKohli Group Title
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    So, \(x^2 - 3x - 4x + 12 = x(x - 3) - 4(x - 3)\). \(x(x - 3) - 4(x - 3) \Longrightarrow ( x - 4)( x - 3)\)

    • one year ago
  30. ParthKohli Group Title
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    Are you there? Do you get it?

    • one year ago
  31. Sunshine447 Group Title
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    yeah

    • one year ago
  32. ParthKohli Group Title
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    Okay, so \((x - 4)(x - 3) = 0\). Do you know the 'zero-product rule'?

    • one year ago
  33. Sunshine447 Group Title
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    i dont think so

    • one year ago
  34. ParthKohli Group Title
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    All right, you have two solutions here. \( \color{Black}{\Rightarrow x - 4 = 0}\) \( \color{Black}{\Rightarrow x - 3 = 0}\)

    • one year ago
  35. Sunshine447 Group Title
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    okay...

    • one year ago
  36. ParthKohli Group Title
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    Can you solve those equations? That is how you get the solutions.

    • one year ago
  37. Sunshine447 Group Title
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    x=4 and 3

    • one year ago
  38. Sunshine447 Group Title
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    Thanks!

    • one year ago
  39. ParthKohli Group Title
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    You're welcome!

    • one year ago
  40. TheViper Group Title
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    \[\LARGE{x^2-7x+12=0}\]\[\Large{x^2-3x-4x+12=0}\]\[\Large{x(x-3)-4(x-3)=0}\]\[\Large{(x-3)(x-4)}\]\[\Large{\color{green}{x=3\space or \space4}}\]

    • one year ago
  41. TheViper Group Title
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    So A. is the right answer ;)

    • one year ago
  42. TheViper Group Title
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    Understood ???

    • one year ago
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