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
4 years, 5 months ago
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
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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.
descartes rule of signs only must a limit on the maximum real roots.i dont think thats the only idea to be applied here.
I do not know about how to prove it has no real roots.
You can show that it has no rational roots.
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
there can be 4 or 2 and maybe 0 positive real roots for the polynomial according to descarte's theorem.
But that is only concluded for negative roots, what about real positive roots?
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..
But f(x) has 4 sign changes.
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
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.
but the question asks us to prove that thr are no real roots.....
and this is nt possible according to descartes theorem
Yeah, that is what I am trying to say.
k i got u this time.....nd r u a bengali?
No, but I live in Kolkata.
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?
Yeah a cbse student.
Are you in the science stream?
ofcourse yes....and r u writing the ISI exam on 12th of may??
Jul 14, 2013
ohh i see....can we talk in gmail if u dont mind
this is my id
send me request
Decartes' rule of signs? Maybe you could use that.