# Find all integer values k for which p(x)=x³-(k+1)x+2k=0 has a rational root.?

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Find all integer values k for which p(x) = x³ -(k+1) x + 2k = 0 has a rational root.

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• 7 years ago

Hello,

x³ – (k + 1)x + 2k = 0

= = = = = = = = = = = = = = = = = = = = = =

Obviously, if k=0, we obtain:

x³ – x = 0

x(x² – 1) = 0

x(x – 1)(x + 1) = 0

which has actually three rational roots.

= = = = = = = = = = = = = = = = = = = = = =

Let's suppose now that k is prime and k≠0:

Then 2k has four divisors: 1, 2, k and 2k.

Rational root theorem induce that if a rational root exists, then it must be in the following set:

{-2k; -k; -2; -1; 1; 2; k; 2k}

►► If -2k is root:

-8k³ + 2k(k + 1) + 2k = 0

-2k(4k² – k – 1 – 1) = 0

4k² – k – 2 = 0 →→→ Since k≠0, we can divide by it.

(2k)² – 2(2k)(¼) + (¼)² – 33/16 = 0 →→→ Complete the square.

(2k – ¼)² – [(√33)/4]² = 0

[2k – ¼ – (√33)/4][2k – ¼ + (√33)/4] = 0

k = (1 ± √33) / 8

and k cannot be an integer; thus -2k cannot be root.

►► If -k is root:

-k³ + k(k + 1) + 2k = 0

-k(k² – k – 1 – 2) = 0

k² – k – 3 = 0 →→→ Since k≠0, we can divide by it.

k² – 2(k)(½) + (½)² – 13/4 = 0 →→→ Complete the square.

(k – ½)² – [(√13)/2]² = 0

[k – ½ – (√13)/2][k – ½ + (√13)/2] = 0

k = (1 ± √13) / 2

and k cannot be an integer; thus -k cannot be root.

►► If -2 is root:

-8 + 2(k + 1) + 2k = 0

-8 + 4k + 2 = 0

4k = 6

k = 3/2

and k is not an integer; thus -2 cannot be root.

►► If -1 is root:

-1 + k + 1 + 2k = 0

3k = 0

k = 0

but k cannot be nil; thus -1 cannot be root.

►► If 1 is root:

1 – k – 1 + 2k = 0

k = 0

but k cannot be nil; thus 1 cannot be root.

►► If 2 is root:

8 – 2(k + 1) + 2k = 0

8 – 2 = 0

which is plain impossible; thus 2 cannot be root.

►► If k is root:

k³ – k(k + 1) + 2k = 0

k(k² – k – 1 + 2) = 0

k² – k + 1 = 0 →→→ Since k≠0, we can divide by it.

k² – 2(k)(½) + (½)² + ¾ = 0 →→→ Complete the square.

(k – ½)² + ¾ = 0

This quadratic equation has no real solution; thus k cannot be root.

►► If 2k is root:

8k³ – 2k(k + 1) + 2k = 0

2k(4k² – k – 1 + 1) = 0

2k²(4k – 1) = 0

4k – 1 = 0 →→→ Since k≠0, we can divide by it.

k = ¼

and k cannot be an integer; thus 2k cannot be root.

►► Thus if k is prime, there cannot be any rational root.

= = = = = = = = = = = = = = = = = = = = = =

If k is not prime and k≠0, then there exists at least an prime integer k' such as k=k'.p

Rational root theorem induces that if a rational root exists, then it must be either in the set

{-2k; -k; -2; -1; 1; 2; k; 2k} or a divisor of p.

By the same earlier logical reasoning above, we deduce there is no rational root in the above set.

And we are left to consider all remaining divisors of p. If p is prime, we proved there was no rational root; otherwise, then there exists at least an prime integer p' such as p=p'.q

And we repeat the process...

►► Thus if k is not prime, there cannot be any rational root.

= = = = = = = = = = = = = = = = = = = = = =

Thus the only integer value for which p(x) would be nil is k=0.

Regards,