find k for 3x^2 + 2kx + 27 =0
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Answer:
NINE AND MINUS NINE
Step-by-step explanation:
Condition of quadratic equation if root is real and equal ,
D=0
b^2-4ac=0
(2k)^2-4*3*27=0
4k^2=4*81
k^2=81
k=+9,-9
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3x^2 + 2kx + 27
A = Coefficient of X² = 3
B = Coefficient of X = 2K
C = Constant term = 27
Now finding discriminent value of required equation and putting it equal to zero as roots will be real and equal*.
b^2-4ac=0
(2k)^2-4*3*27=0
4k^2=4*81
k^2=81
k=+9,-9
Hence +9 and -9 are required values of K.
A = Coefficient of X² = 3
B = Coefficient of X = 2K
C = Constant term = 27
Now finding discriminent value of required equation and putting it equal to zero as roots will be real and equal*.
b^2-4ac=0
(2k)^2-4*3*27=0
4k^2=4*81
k^2=81
k=+9,-9
Hence +9 and -9 are required values of K.
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