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

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Comments

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

Gabriel Wong - 8 years, 1 month 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.

Gabriel Wong - 8 years, 1 month ago

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

Aditya Parson - 8 years, 1 month 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.

pranav chakravarthy - 8 years, 1 month 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

Mishti Angel - 8 years, 1 month ago

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

Aditya Parson - 8 years, 1 month 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..

Mishti Angel - 8 years, 1 month ago

@Mishti Angel – But f(x) has 4 sign changes.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – 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

Mishti Angel - 8 years, 1 month ago

@Mishti Angel – 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.

Aditya Parson - 8 years, 1 month ago

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

Mishti Angel - 8 years, 1 month ago

@Mishti Angel – Yeah, that is what I am trying to say.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – k i got u this time.....nd r u a bengali?

Mishti Angel - 8 years, 1 month ago

@Mishti Angel – No, but I live in Kolkata.

Aditya Parson - 8 years, 1 month ago

@Aditya Parson – 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?

Mishti Angel - 8 years, 1 month ago

@Mishti Angel – Yeah a cbse student. Are you in the science stream?

Aditya Parson - 8 years, 1 month ago

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

Mishti Angel - 8 years, 1 month ago

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

Aditya Parson - 8 years, 1 month ago

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

Ameer Lubang - 8 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$