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

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

No vote yet

2 votes

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestLet 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 – Gabriel Wong · 3 years, 10 months ago

Log in to reply

Thus x^8 - x^7 + x^2 - x + 15 is always positive and thus has no real roots. – Gabriel Wong · 3 years, 10 months ago

Log in to reply

descartes rule of signs only must a limit on the maximum real roots.i dont think thats the only idea to be applied here. – Pranav Chakravarthy · 3 years, 10 months ago

Log in to reply

I do not know about how to prove it has no real roots. You can show that it has no rational roots. – Aditya Parson · 3 years, 10 months ago

Log in to reply

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 – Mishti Angel · 3 years, 10 months ago

Log in to reply

– Aditya Parson · 3 years, 10 months ago

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

– Aditya Parson · 3 years, 10 months ago

But that is only concluded for negative roots, what about real positive roots?Log in to reply

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.. – Mishti Angel · 3 years, 10 months ago

Log in to reply

– Aditya Parson · 3 years, 10 months ago

But f(x) has 4 sign changes.Log in to reply

so wht can u conclude from this?.... i guess thr is a flaw in the question – Mishti Angel · 3 years, 10 months ago

Log in to reply

– Aditya Parson · 3 years, 10 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.Log in to reply

– Mishti Angel · 3 years, 10 months ago

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

– Aditya Parson · 3 years, 10 months ago

Yeah, that is what I am trying to say.Log in to reply

– Mishti Angel · 3 years, 10 months ago

k i got u this time.....nd r u a bengali?Log in to reply

– Aditya Parson · 3 years, 10 months ago

No, but I live in Kolkata.Log in to reply

so u've finished ur 12th this yr?...r u a cbse student? – Mishti Angel · 3 years, 10 months ago

Log in to reply

– Aditya Parson · 3 years, 10 months ago

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

– Mishti Angel · 3 years, 10 months ago

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

Log in to reply

– Mishti Angel · 3 years, 10 months ago

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

Decartes' rule of signs? Maybe you could use that. – Ameer Lubang · 3 years, 10 months ago

Log in to reply