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Prove the equation has no real solution

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
Let $f(x) = x^8-x^7+x^2-x = (x-1)(x^7 + x)$. If $x<0$ then both factors are negative and so $f(x)>0$. If $x>1$ then both factors are positive and so $f(x)>0$ again. If $0\leqslant x \leqslant 1$ then $|x-1|\leqslant 1$ and $|x^7+x|\leqslant 2$ and so $f(x) \geqslant -2$. Thus $f(x) \geqslant -2$ for all real $x$ and so $f(x) + 15 \geqslant 13 >0$. Hence $f(x) + 15 = 0$ has no real roots.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,690
Let $f(x) = x^8-x^7+x^2-x = (x-1)(x^7 + x)$. If $x<0$ then both factors are negative and so $f(x)>0$. If $x>1$ then both factors are positive and so $f(x)>0$ again. If $0\leqslant x \leqslant 1$ then $|x-1|\leqslant 1$ and $|x^7+x|\leqslant 2$ and so $f(x) \geqslant -2$. Thus $f(x) \geqslant -2$ for all real $x$ and so $f(x) + 15 \geqslant 13 >0$. Hence $f(x) + 15 = 0$ has no real roots.
Thanks for participating, Opalg and welcome back to the forum!!! You have not been posting much lately and we have certainly missed you.(Sun)
 

mente oscura

Well-known member
Nov 29, 2013
172
Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.
Hello.


[tex]y=x^8-x^7+x^2-x+15[/tex]

[tex]y'=8x^7-7x^6+2x-1=0[/tex]

[tex]x_0 \approx{ }0.53079[/tex](unique real solution)

[tex]y''=56x^6-42x^5+2[/tex]

[tex]For \ x_0 \approx{ }0.53079 \rightarrow{ }y'' \approx{ }1.48279 \rightarrow{ }x_0 \ it's \ minimum[/tex]

[tex]y_0 \approx{ }14.7453 \rightarrow{}y \cancel{=} \ 0 \ (never)[/tex]



Regards.
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Hello.


[tex]y=x^8-x^7+x^2-x+15[/tex]

[tex]y'=8x^7-7x^6+2x-1=0[/tex]

[tex]x_0 \approx{ }0.53079[/tex](unique real solution)

[tex]y''=56x^6-42x^5+2[/tex]

[tex]For \ x_0 \approx{ }0.53079 \rightarrow{ }y'' \approx{ }1.48279 \rightarrow{ }x_0 \ it's \ minimum[/tex]

[tex]y_0 \approx{ }14.7453 \rightarrow{}y \cancel{=} \ 0 \ (never)[/tex]



Regards.
Hey mente oscura, thanks for participating and yes, your approach works as well! Well done!:)
 

jacks

Well-known member
Apr 5, 2012
226
$\bf{Solution::}$ Let $f(x) = x^8-x^7+x^2-x+15$, we check real solution in $\bf{x\in \mathbb{R}}$

$\bullet$ If $x\leq 0$, Then $f(x) = x^8-x^7+x^2-x+15>0$

$\bullet$ If $0<x<1$, Then $f(x) = x^8+x^2(1-x^5)+(1-x)+14>0$

$\bullet$ If $x\geq 1$, Then $f(x)= x^7(x-1)+x(x-1)+15>0$

So we observe that $f(x)= x^8-x^7+x^2-x+15>0\;\forall x\in \mathbb{R}$

So $f(x)=x^8-x^7+x^2-x+1=0$ has no real roots for all $x\in \mathbb{R}$

- - - Updated - - -

**Another Solution::**

If $x<0,$ note that $x^8+(-x^7)+x^2+(-x)>0,$ so the polynomial cannot have any negative roots.


If $x\geq 0,$ then note that from AM-GM inequality we have:
$\left\{\begin{aligned}& \frac 78 x^8+\frac 18\geq x^7\\& x^2+\frac 14\geq x\end{aligned}\right\} ;$

Thus $\displaystyle\frac 78x^8-x^7+x^2-x+\frac 38>0;$
 
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anemone

MHB POTW Director
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Feb 14, 2012
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$\bf{Solution::}$ Let $f(x) = x^8-x^7+x^2-x+15$, we check real solution in $\bf{x\in \mathbb{R}}$

$\bullet$ If $x\leq 0$, Then $f(x) = x^8-x^7+x^2-x+15>0$

$\bullet$ If $0<x<1$, Then $f(x) = x^8+x^2(1-x^5)+(1-x)+14>0$

$\bullet$ If $x\geq 1$, Then $f(x)= x^7(x-1)+x(x-1)+15>0$

So we observe that $f(x)= x^8-x^7+x^2-x+15>0\;\forall x\in \mathbb{R}$

So $f(x)=x^8-x^7+x^2-x+1=0$ has no real roots for all $x\in \mathbb{R}$

- - - Updated - - -

**Another Solution::**

If $x<0,$ note that $x^8+(-x^7)+x^2+(-x)>0,$ so the polynomial cannot have any negative roots.


If $x\geq 0,$ then note that from AM-GM inequality we have:
$\left\{\begin{aligned}& \frac 78 x^8+\frac 18\geq x^7\\& x^2+\frac 14\geq x\end{aligned}\right\} ;$

Thus $\displaystyle\frac 78x^8-x^7+x^2-x+\frac 38>0;$
Hey jacks, thanks for participating and it's so nice to see so many people answered to my challenge problem!:eek: