find the value of k in quartic equation 9xsquare-kx+16 =0​

Question

find the value of k in quartic equation 9xsquare-kx+16 =0​

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Alexandra 1 month 2021-08-17T17:23:38+00:00 2 Answers 0 views 0

Answers ( )

    0
    2021-08-17T17:24:51+00:00

    Answer:

    k is +3/-3…….

    hope it helps

    0
    2021-08-17T17:24:56+00:00

    Answer:

    The value of determinant should be zero.

    i.e. D=b²-4ac=0———(1)

    Where in this case,

    b=8k

    a=9 , and

    c=16

    Substituting the above values in (1)

    =>64k²-4*9*16=0

    =>64k²-576=0

    =>64k²=576

    =>k²=576/64

    =>k²=9

    =>k=+3 or -3

    Hence , for the equation nto have equal roots the value of k is either +3 or -3

    Step-by-step explanation:

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