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anurag singhal Grade:
        

no. of real roots of equation is equal to


x8-x5+x2-x+1=0


a) 0


b) 2


c) 4


d) 6

8 years ago

Answers : (2)

Askiitian.Expert Rajat
24 Points
										

4363-239_6001_real roots.JPG

8 years ago
mycroft holmes
266 Points
										

Let f(x) = x8-x5+x2-x+1


 


f(0) = 1 = f(1)


 


Its obvious that for x<0, f(x)>0


 


For x>1 we have x8>x5 and x2>x so that f(x)>1


 


Again for 0<x<1, we write f(x) = x8 +(x2-x5) + (1-x) and we know that in this domain we have (x2>x5) and (1>x).


 


So again f(x)>0


 


Thus f(x)>0 for all real x and hence this polynomial has no real roots


 

8 years ago
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