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    need help in this question
    i tried a couple of time and still got the same answer
    ( will upload the question)
    so there is quadratic equation question
    the thing that i got wrong is that when i solved this by factorization
    i got 1 and sqroot 3=1.73.....
    and when i solve by quadratic formula i got
    1 and 0.73

    can someone try please
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    (Original post by Qer)
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    Your quadratic is in \tan\theta

    So, what's the coefficient with \tan^2\theta?

    It's 1, so with the standard a,b,c, we have a=1

    But what about the coefficient for \tan\theta?

    That's -(1+\sqrt{3}) and there's your b.

    And c, the constant term = ...?
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    (Original post by ghostwalker)
    Your quadratic is in \tan\theta

    So, what's the coefficient with \tan^2\theta?

    It's 1, so with the standard a,b,c, we have a=1

    But what about the coefficient for \tan\theta?

    That's -(1+\sqrt{3}) and there's your b.

    And c, the constant term = ...?
    i solve like that
    sec^2 -(1+3) tan+sq(3)=1
    1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
    tan^2 -sqrooot 3 tan + sqroot 3=1
    tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

    so from here
    a=1
    b= - sqroot 3
    c=(sqroot 3 -1)
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    (Original post by Qer)
    i solve like that
    sec^2 -(1+3) tan+sq(3)=1
    1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
    tan^2 -sqrooot 3 tan + sqroot 3=1
    tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

    so from here
    a=1
    b= - sqroot 3
    c=(sqroot 3 -1)
    How did you go from your second to third line?
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    (Original post by Zacken)
    How did you go from your second to third line?
    1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1 ( second )
    tan^2 -sqrooot 3 tan + sqroot 3=1 ( third)



    expand bracket
    1 + tan^2 -1 -sqroot 3 tan +sqroot 3=1 ( i cancel +1 and -1 here)
    that give me that.................
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    (Original post by ghostwalker)
    Your quadratic is in \tan\theta

    So, what's the coefficient with \tan^2\theta?

    It's 1, so with the standard a,b,c, we have a=1

    But what about the coefficient for \tan\theta?

    That's -(1+\sqrt{3}) and there's your b.

    And c, the constant term = ...?
    c= sqroot 3-1?

    but still not getting right answer
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    (Original post by Qer)
    1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1 ( second )
    tan^2 -sqrooot 3 tan + sqroot 3=1 ( third)



    expand bracket
    1 + tan^2 -1 -sqroot 3 tan +sqroot 3=1 ( i cancel +1 and -1 here)
    that give me that.................
    -(1+sq(3))tan = -tan - sq(3) tan


    Not

    -(1+sqrt(3))tan = -1 - sqrt(3) tan
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    (Original post by Zacken)
    -(1+sq(3))tan = -tan - sq(3) tan


    Not

    -(1+sqrt(3))tan = -1 - sqrt(3) tan
    yeah i got it after replying you

    a =1
    b=-1-sqroot 3
    c=sqrt3 - 1

    -(-1-sqroot3) +- SQRT ( -1-sqroot3)^2 -4(a)(sq root 3 -1) / 2

    not getting the right answer
    can you please try this for me
    thanks
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    (Original post by Qer)
    yeah i got it after replying you

    a =1
    b=-1-sqroot 3
    c=sqrt3 - 1

    -(-1-sqroot3) +- SQRT ( -1-sqroot3)^2 -4(a)(sq root 3 -1) / 2

    not getting the right answer
    can you please try this for me
    thanks
    Huh?

    From the question and the second implication line, you have:

    a=1
    b=-(1+\sqrt{3})
    c=\sqrt{3}

    Why are you saying that c=\sqrt{3}-1...?
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    (Original post by Qer)
    i solve like that
    sec^2 -(1+3) tan+sq(3)=1
    1 + tan^2 -(1+ sqroot (3) ) tan +sqroot 3 =1
    tan^2 -sqrooot 3 tan + sqroot 3=1
    tan^2 - sqroot 3 tan + (sqroot 3 -1)=0

    so from here
    a=1
    b= - sqroot 3
    c=(sqroot 3 -1)
    When you change sec go tan, you can cancel 1 which appears on both sides.

    This gives

    a = 1

    b = -(√3+1)

    c = √3
 
 
 
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