find the value of k in quartic equation 9xsquare-kx+16 =0
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find the value of k in quartic equation 9xsquare-kx+16 =0
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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:
Answer:
k is +3/-3.......
hope it helps