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# No real roots

Show that the polynomial x^8 -x^7+x^2 -x+15=0 Has no real roots.

Need this for an exam Urgently any kind of help would be appreciated

Note by Pranav Chakravarthy
4 years, 8 months ago

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Let K = x^8-x^7 + x^2 - x = x(x-1)(x^6+1)

Observe that x must be in (0,1) for K to be negative

x^6+1 takes values in (1,2) for x in the range (0,1)

The minimum value of (x)(x-1) is -1/4

Thus K >= -1/2

- 4 years, 8 months ago

From the above, K + 15 >= 29/2

Thus x^8 - x^7 + x^2 - x + 15 is always positive and thus has no real roots.

- 4 years, 8 months ago

descartes rule of signs only must a limit on the maximum real roots.i dont think thats the only idea to be applied here.

- 4 years, 8 months ago

I do not know about how to prove it has no real roots. You can show that it has no rational roots.

- 4 years, 8 months ago

actually if we are given with a polynomial where f(x) = 0 ., then for f(x) to have no real roots, the condition is that f(-x) should not have ny change of signs...... so in the given polynomial f(-x) = (-x)^8 -(-x)^7 +(-x)^2 -(-x)+15 => f(-x) = x^8 +x^7+x^2+x+15
so thr is no change of signs in f(-x)....and hence there r no real roots , all of the roots r imaginary...

here i've used the descartes rule of signs

- 4 years, 8 months ago

there can be 4 or 2 and maybe 0 positive real roots for the polynomial according to descarte's theorem.

- 4 years, 8 months ago

But that is only concluded for negative roots, what about real positive roots?

- 4 years, 8 months ago

no thts not correct actually......the maximum number of negative real roots of a polynomial equation f(x)=0 is the number of change of signs from positive to negative and negative to positive in f(-x)

and the number of positive real roots of a polynomial equation f(x)=0 is the number of change of signs from positive to negative and negative to positive in f(x)

and if f(-x) has got no sign change, then the polynomial equation has complex roots..

- 4 years, 8 months ago

But f(x) has 4 sign changes.

- 4 years, 8 months ago

yaa tht is also true....so atmost it can have 4 positive real roots.....but f(-x) has no sign change ...

so wht can u conclude from this?.... i guess thr is a flaw in the question

- 4 years, 8 months ago

Read Descartes theorem. It states that if f(x) has sign changes for eg 4 then it has 4 positive real roots or 2 and then maybe 0. If f(-x) has sign changes then there are for eg some negative real roots again. Here, there are no negative roots for sure, but the possibility of positive real roots is 4,2 or 0.

- 4 years, 8 months ago

but the question asks us to prove that thr are no real roots..... and this is nt possible according to descartes theorem

- 4 years, 8 months ago

Yeah, that is what I am trying to say.

- 4 years, 8 months ago

k i got u this time.....nd r u a bengali?

- 4 years, 8 months ago

No, but I live in Kolkata.

- 4 years, 8 months ago

ohh .... i m a bong....nd its a pleasure to meet u

so u've finished ur 12th this yr?...r u a cbse student?

- 4 years, 8 months ago

Yeah a cbse student. Are you in the science stream?

- 4 years, 8 months ago

ofcourse yes....and r u writing the ISI exam on 12th of may??

- 4 years, 8 months ago

Comment deleted Jul 14, 2013

ohh i see....can we talk in gmail if u dont mind this is my id send me request prerana.chakrabarti@gmail.com

- 4 years, 8 months ago

Decartes' rule of signs? Maybe you could use that.

- 4 years, 8 months ago