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koli123able

  • 2 years ago

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

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  1. koli123able
    • 2 years ago
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    how should i start

  2. myininaya
    • 2 years ago
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    For a root with multiplicity two you need to need to have the discriminant equal to 0.

  3. koli123able
    • 2 years ago
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    D=b^2-4ac

  4. myininaya
    • 2 years ago
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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

  5. koli123able
    • 2 years ago
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    but what is b and a and c the equation is a bit confusing

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

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

  8. myininaya
    • 2 years ago
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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

  9. koli123able
    • 2 years ago
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    \[x^2+(7+a)x+7a+1=0\]

  10. koli123able
    • 2 years ago
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    A=1 B=(7+a) c=(7a+1)

  11. myininaya
    • 2 years ago
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    Yep! :)

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

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

  14. koli123able
    • 2 years ago
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    ok wait

  15. koli123able
    • 2 years ago
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    49+14a+a^2-14a+4=0

  16. myininaya
    • 2 years ago
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    I think you mean -28a since -4(7)=-28 not -14

  17. koli123able
    • 2 years ago
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    yeah -28a

  18. koli123able
    • 2 years ago
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    a^2-14a+53=0

  19. koli123able
    • 2 years ago
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    is it right

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

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

  22. koli123able
    • 2 years ago
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    a^2-14a+45=0

  23. myininaya
    • 2 years ago
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    yep do you know how to factor?

  24. koli123able
    • 2 years ago
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    yep but i am solving this with quadratic is it ok?

  25. myininaya
    • 2 years ago
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    The quadratic formula? That is fine. :)

  26. koli123able
    • 2 years ago
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    yes

  27. Miyuru
    • 2 years ago
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    I solved it by factors..

  28. Miyuru
    • 2 years ago
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    thnx @myininaya U are great. Nice help

  29. koli123able
    • 2 years ago
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    thanks

  30. koli123able
    • 2 years ago
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    got 2 values 9 and 5 of a

  31. myininaya
    • 2 years ago
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    Great! You go guys! :)

  32. myininaya
    • 2 years ago
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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.

  33. myininaya
    • 2 years ago
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    The discriminant is \[B^2-4AC \]

  34. Miyuru
    • 2 years ago
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    How did it came.

  35. Miyuru
    • 2 years ago
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    @myininaya how to solve the equation to get the discriminant..

  36. myininaya
    • 2 years ago
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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.

  37. Miyuru
    • 2 years ago
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    So to slove the equation should i find a b c

  38. Miyuru
    • 2 years ago
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    *solve

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