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no. of real roots of equation is equal to x 8 -x 5 +x 2 -x+1=0 a) 0 b) 2 c) 4 d) 6

no. of real roots of equation is equal to


x8-x5+x2-x+1=0


a) 0


b) 2


c) 4


d) 6

Grade:

2 Answers

Askiitian.Expert Rajat
24 Points
12 years ago

4363-239_6001_real roots.JPG

mycroft holmes
272 Points
12 years ago

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

 

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