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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
```
```
8 years ago
mycroft holmes
271 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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