Show all real roots are negative

In summary, we need to show that all real roots of the polynomial $f(x) = x^5 - 10x + 38$ are negative. This can be done by rewriting the function as $f(x) = x(x^4 - 10) + 38$ and considering the domain when $x \geq 10^{\tiny\dfrac{1}{4}}$. Apologies for the previous confusion.
  • #1
anemone
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Show that all real roots of the polynomial $f(x)=x^5-10x+38$ are negative.

Note:

I know this is a fairly easy challenge, but it's good to see how different approaches can be generated from different people so that we can learn from one another. :eek: (Yes)
 
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  • #2
anemone said:
Show that all real roots of the polynomial $f(x)=x^5-10x+38$ are negative.

Note:

I know this is a fairly easy challenge, but it's good to see how different approaches can be generated from different people so that we can learn from one another. :eek: (Yes)

$f(x)= x^5 + 10(3.8-x) $
it is >0 for $0\le x\lt3.8$

further
$f(x) = x(x^4-10) + 38$

for $x\gt 2$ above is > 0 as $2^4 = 16 \gt 10$

so above is >0 for $2\le x$

we have shown that it is positive for x > 0 so no root 0 or positive hence all real roots are -ve.
 
Last edited:
  • #3
kaliprasad said:
$f(x)= x^5 + 10(3.8-x) $
it is >0 for $0\le x\lt3.8$

further
$f(x) = x(x^4-10) + 38$

for $x\gt 2$ above is > 0 as $2^4 = 16 \gt 10$

so above is >0 for $2\le x$

we have shown that it is positive for x > 0 so no root 0 or positive hence all real roots are -ve.

Hi kaliprasad,

Thanks for participating.:) I am curious, is there any chance when you rewritten the function of $f$ as $f(x) = x(x^4-10) + 38$, you mean to imply the domain when $x\ge 10^{\tiny\dfrac{1}{4}}$?
 
  • #4
anemone said:
Hi kaliprasad,

Thanks for participating.:) I am curious, is there any chance when you rewritten the function of $f$ as $f(x) = x(x^4-10) + 38$, you mean to imply the domain when $x\ge 10^{\tiny\dfrac{1}{4}}$?

You are right but I have shown that it is true for x < 3.8 from the 1st equation and I have chosen a suitable value < 3.8 ( that is 2) to show that it is true for x > 2. $x\ge 10^{\tiny\dfrac{1}{4}}$ condition is a superset of it. but if it is true for x > 2 it does meet the criteria
 
  • #5
kaliprasad said:
You are right but I have shown that it is true for x < 3.8 from the 1st equation and I have chosen a suitable value < 3.8 ( that is 2) to show that it is true for x > 2. $x\ge 10^{\tiny\dfrac{1}{4}}$ condition is a superset of it. but if it is true for x > 2 it does meet the criteria

Oh My...I don't know what is wrong with me! How could I ask something so stupid! Sorry kali, I so wish to retract what I have asked just to appeared to be less silly...:eek:
 

Related to Show all real roots are negative

What does it mean when a polynomial has all real roots that are negative?

When a polynomial has all real roots that are negative, it means that all of the solutions to the polynomial equation are negative numbers. This means that when the polynomial is set equal to zero, all of the resulting solutions will be negative numbers.

How do you determine if a polynomial has all real roots that are negative?

To determine if a polynomial has all real roots that are negative, you can graph the polynomial and see if all of the x-intercepts are located in the negative portion of the graph. Another way is to use the Descartes' Rule of Signs to count the number of positive and negative roots.

Can a polynomial have both positive and negative real roots?

Yes, a polynomial can have both positive and negative real roots. This means that some of the solutions to the polynomial equation will be positive numbers and some will be negative numbers.

Do complex roots count as real roots when determining if they are negative?

No, complex roots do not count as real roots when determining if they are negative. Complex roots are solutions to the polynomial equation that involve imaginary numbers, whereas real roots are solutions that only involve real numbers.

How does the degree of a polynomial affect the number of negative real roots?

The degree of a polynomial can affect the number of negative real roots. For a polynomial with an odd degree, there will always be at least one negative real root. However, for a polynomial with an even degree, there may be zero, two, or more negative real roots.

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