If the eq. x^-bx+1=0 does not posses real roots,then find the number in which b lies
tiasaghosh5876:
any digit for x????????????
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I think you wanted to say x²-bx+1.....
As you said that it doesn't posses real root
So b²-4ac must be less than 0.
Here a and c both equals 1.
b²<4 that way.
b<±4 which has two conditions
The one satisfying is b<-4.
As you said that it doesn't posses real root
So b²-4ac must be less than 0.
Here a and c both equals 1.
b²<4 that way.
b<±4 which has two conditions
The one satisfying is b<-4.
yes.. but i can't solve it
See , if an equation ax²+bx+c doesn't have a real root that means b²-4ac < 0.
YEAAH
SO WHAT IS THE ANSWER
The range for b is less than -4.or you can say b belongs to (-infinity,-4].
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