MHB Prove the equation has no real solution

[anemone]

Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.

 

[Opalg]

[sp]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.[/sp]

 

[anemone]

[sp]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.[/sp]

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]

Prove that the polynomial equation $x^8-x^7+x^2-x+15=0$ has no real solution.

Hello.

Spoiler

y=x^8-x^7+x^2-x+15

y&#039;=8x^7-7x^6+2x-1=0

x_0 \approx{ }0.53079(unique real solution)

y&#039;&#039;=56x^6-42x^5+2

For \ x_0 \approx{ }0.53079 \rightarrow{ }y&#039;&#039; \approx{ }1.48279 \rightarrow{ }x_0 \ it&#039;s \ minimum

y_0 \approx{ }14.7453 \rightarrow{}y \cancel{=} \ 0 \ (never)

Regards.

 

[anemone]

Hello.

Spoiler

y=x^8-x^7+x^2-x+15

y&#039;=8x^7-7x^6+2x-1=0

x_0 \approx{ }0.53079(unique real solution)

y&#039;&#039;=56x^6-42x^5+2

For \ x_0 \approx{ }0.53079 \rightarrow{ }y&#039;&#039; \approx{ }1.48279 \rightarrow{ }x_0 \ it&#039;s \ minimum

y_0 \approx{ }14.7453 \rightarrow{}y \cancel{=} \ 0 \ (never)

Regards.

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

 

[juantheron]

Spoiler

$\bf{Solution::}$ 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::**

Spoiler

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

 

[anemone]

Spoiler

$\bf{Solution::}$ 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::**

Spoiler

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!