Given f(x) = x^3 + (1 - k^2)x + k, show -k is root of f, find other 2 roots, find....

[markosheehan]

f (x) = x³ + (1 - k²) x + k is a cubic where k is a constant.

1. Show that -k is a root of f.

2. Find, in terms of k, the other two roots of f.

3. Find the set of values for k for which f has exactly one real root.

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i think i know how to do the first question just sub in k for x and it should equal 0 but i dont know how to do the second and third qustion.

 

[stapel]

f (x) = x³ + (1 - k²) x + k is a cubic where k is a constant.

1. Show that -k is a root of f.

2. Find, in terms of k, the other two roots of f.

3. Find the set of values for k for which f has exactly one real root.

---

i think i know how to do the first question just sub in k for x and it should equal 0 but i dont know how to do the second and third qustion.

If the value is a root, then what must be a factor? If this factor is divided out (leaving a quadratic), how then may the other two roots be found?

 

[markosheehan]

yes denis that is the qustion and i have worked out the first part. i am stuck on the second part you have to divide x³+(1-k²)x+k by x+k i am finding this very hard to do can someone do it for me

 

[Bob Brown MSEE]

hint

Create Conjectures, then work backward:
1 Posit: r1+r2 = k
where r1, r2 are the other 2 roots
2 Posit: r1=r2=1 when k=2
3 Verify, then work backward to get r1(k) and r2(k) proofs

 

[stapel]

yes denis that is the qustion and i have worked out the first part. i am stuck on the second part you have to divide x³+(1-k²)x+k by x+k i am finding this very hard to do can someone do it for me

We won't be taking your tests for you, and we're not (supposed to be) doing your homework for you. You need to learn the material they've covered in class.

To learn how to divide polynomials, try here. Once you have studied at least two lessons from the listing, please attempt the long division. If you get stuck, please then show your work so far, so we can help you get un-stuck.

Thank you!

 

[Deleted member 4993]

f (x) = x³ + (1 - k²) x + k is a cubic where k is a constant.

1. Show that -k is a root of f.

2. Find, in terms of k, the other two roots of f.

3. Find the set of values for k for which f has exactly one real root.

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i think i know how to do the first question just sub in k for x and it should equal 0 but i dont know how to do the second and third qustion.

Just to start you off:

f (x) = x³ + (1 - k²) x + k

f (x) = x³ - k² x + x + k

Factorize the Right-hand-side (RHS) and continue....

 

[pka]

f (x) = x³ + (1 - k²) x + k is a cubic where k is a constant.
2. Find, in terms of k, the other two roots of f.
3. Find the set of values for k for which f has exactly one real root.

\(\displaystyle \dfrac{x^3+(1-k^2)x+k}{x+k}=x^2-kx+1\), the other roots: \(\displaystyle \dfrac{k\pm\sqrt{k^2-4}}{2}\).

What if \(\displaystyle |k|<2~?\)

 

[stapel]

http://www.sosmath.com/CBB/viewtopic.php?f=1&t=68381

At how many sites did you post this?

Enough to get all the steps done for him!