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koli123able Group Title

If the equation x^2+(7+a)x+7a+1=0 has equal roots,find the value of a. . . . ANSWER = {9,5}

  • one year ago
  • one year ago

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  1. koli123able Group Title
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    how should i start

    • one year ago
  2. myininaya Group Title
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    For a root with multiplicity two you need to need to have the discriminant equal to 0.

    • one year ago
  3. koli123able Group Title
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    D=b^2-4ac

    • one year ago
  4. myininaya Group Title
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    So what I'm saying is: \[Ax^2+Bx+C=0 => \text{ discriminant } = B^2-4AC\] And we want here for the discriminant=0

    • one year ago
  5. koli123able Group Title
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    but what is b and a and c the equation is a bit confusing

    • one year ago
  6. myininaya Group Title
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    Since we want root with multiplicity 2 :) So \[B^2-4AC=0\]

    • one year ago
  7. myininaya Group Title
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    \[\text{ you have } x^2+(7+a)x+(7a+1)=0\]

    • one year ago
  8. myininaya Group Title
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    A is the number in front of x^2 B is the number in front of x C is the leftover part

    • one year ago
  9. koli123able Group Title
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    \[x^2+(7+a)x+7a+1=0\]

    • one year ago
  10. koli123able Group Title
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    A=1 B=(7+a) c=(7a+1)

    • one year ago
  11. myininaya Group Title
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    Yep! :)

    • one year ago
  12. myininaya Group Title
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    \[(7+a)^2-4(1)(7a+1)=0\]

    • one year ago
  13. myininaya Group Title
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    That is B^2-4AC=0 You think you can solve that reply above for a?

    • one year ago
  14. koli123able Group Title
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    ok wait

    • one year ago
  15. koli123able Group Title
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    49+14a+a^2-14a+4=0

    • one year ago
  16. myininaya Group Title
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    I think you mean -28a since -4(7)=-28 not -14

    • one year ago
  17. koli123able Group Title
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    yeah -28a

    • one year ago
  18. koli123able Group Title
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    a^2-14a+53=0

    • one year ago
  19. koli123able Group Title
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    is it right

    • one year ago
  20. myininaya Group Title
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    well .... I think your 53 is a little off. you have 49-4 you forgot to distribute that - earlier

    • one year ago
  21. myininaya Group Title
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    \[49+14a+a^2-4(7a+1)=0\] \[49+14a+a^2-28a-4=0\] Now try combining like terms.

    • one year ago
  22. koli123able Group Title
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    a^2-14a+45=0

    • one year ago
  23. myininaya Group Title
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    yep do you know how to factor?

    • one year ago
  24. koli123able Group Title
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    yep but i am solving this with quadratic is it ok?

    • one year ago
  25. myininaya Group Title
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    The quadratic formula? That is fine. :)

    • one year ago
  26. koli123able Group Title
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    yes

    • one year ago
  27. Miyuru Group Title
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    I solved it by factors..

    • one year ago
  28. Miyuru Group Title
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    thnx @myininaya U are great. Nice help

    • one year ago
  29. koli123able Group Title
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    thanks

    • one year ago
  30. koli123able Group Title
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    got 2 values 9 and 5 of a

    • one year ago
  31. myininaya Group Title
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    Great! You go guys! :)

    • one year ago
  32. myininaya Group Title
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    If you guys didn't understand why I set the discriminant equal to 0, then I will try to explain as best as possible. If discriminant is equal to 0, then you will have on root (multiplicity 2) If discriminant is equal to negative number, you have two imaginary solutions If discriminant is equal to positive number, you have two real solutions.

    • one year ago
  33. myininaya Group Title
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    The discriminant is \[B^2-4AC \]

    • one year ago
  34. Miyuru Group Title
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    How did it came.

    • one year ago
  35. Miyuru Group Title
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    @myininaya how to solve the equation to get the discriminant..

    • one year ago
  36. myininaya Group Title
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    \[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a} \text{ is called the quadratic formual } \] The discriminant is that thing under the radical It is what determines if you have one root (w/ multiplicity 2) , imaginary roots, or real roots.

    • one year ago
  37. Miyuru Group Title
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    So to slove the equation should i find a b c

    • one year ago
  38. Miyuru Group Title
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    *solve

    • one year ago
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