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    Can any one show me how to find the two roots in (4x^2 +9)?
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    (Original post by SakuraE)
    Can any one show me how to find the two roots in (4x^2 +9)?
    Are you familiar with using the discriminant to determine the nature of the roots of a function? b^2 - 4ac?
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    (Original post by SakuraE)
    Can any one show me how to find the two roots in (4x^2 +9)?
    4x^2+9=0

    4x^2=-9
    x^2=-9/4
    x=sqrt of -9/4

    x=+3/2i and -3/2i
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    (Original post by maruchan)
    4x^2+9=0

    4x^2=-9
    x^2=-9/4
    x=sqrt of -9/4

    x=+3/2i and -3/2i
    Assumed they have covered complex roots!
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    (Original post by SakuraE)
    Can any one show me how to find the two roots in (4x^2 +9)?
    4x^2 = - 9 \Rightarrow x^2 = -\frac{9}{4} \Rightarrow x = \pm i \sqrt{\frac{9}{4}}
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    (Original post by offhegoes)
    Assumed they have covered complex roots!
    yes!
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    (Original post by offhegoes)
    Assumed they have covered complex roots!
    Which is in the syllabus for FP1, yes.
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    Thank you I get it now I put it through the quadratic formula and it came out as +and - 3/4i
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    (Original post by SakuraE)
    Thank you I get it now I put it through the quadratic formula and it came out as +and - 3/4i
    That doesn't look quite right.
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    (Original post by Zacken)
    That doesn't look quite right.
    after looking through all my workings it seems that as there was two 4s under the radical sign I accidentally missed one when I multiplied it out
 
 
 
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