no. of real roots of equation is equal to

x⁸-x⁵+x²-x+1=0

a) 0

b) 2

c) 4

d) 6

Question

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

Askiitian.Expert Rajat · Answer

mycroft holmes · Answer

Let f(x) = x 8 -x 5 +x 2 -x+1 f(0) = 1 = f(1) Its obvious that for x<0, f(x)>0 For x>1 we have x 8 >x 5 and x 2 >x so that f(x)>1 Again for 0x 5 ) 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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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

2 Answers

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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

2 Answers

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no. of real roots of equation is equal to

x⁸-x⁵+x²-x+1=0

a) 0

b) 2

c) 4

d) 6