Finding $|k|$ of the Polynomial $x^3-kx+25$

[anemone]

The polynomial $x^3-kx+25$ has three real roots. Two of these root sum to 5. What is $|k|$?

 

[topsquark]

[sp]
Let the roots be a, b, and c.

Then by Vieta's formula
a + b + c = 0

Take b + c = 5. Then a = -5,

So we know that
\(\displaystyle a^3 - k a + 25 = 0\)

\(\displaystyle (-5)^3 - k ( -5 ) + 25 = 0\)

k = 20

Note that we haven't used the condition that all roots are real, so let's assume that b and c are complex and that this gives a contradiction.

We know that a = -5 and that b and c are, by assumption, complex. So let b = m + in and c = 5 - (m + in).

The other Vieta formula says that
abc = 25

\(\displaystyle (-5) ( m + in) ( 5 - (m + in) ) = 25\)

\(\displaystyle (-m^2 + 5m + n^2) + (5n - 2mn)i = -5\)

Using the second term
\(\displaystyle 5n - 2mn = 0\)

So n = 0 or m = 5/2.

Let m = 5/2 and put it into the first term:
\(\displaystyle -m^2 + 5m + n^2 = -5\)

\(\displaystyle - \left ( \dfrac{5}{2} \right ) ^2 + 5 \left ( \dfrac{5}{2} \right ) + n^2 = -5\)

\(\displaystyle n^2 = -\dfrac{45}{4}\)
which says that \(\displaystyle n^2 < 0\), which is impossible.

Thus n = 0 and the solution for k = 20 stands as the only possible k.
[/sp]
-Dan

 

[kaliprasad]

Above is good method and here is my

Spoiler: solution

as $x^2$ term is zero and sum of 2 roots is 5 so $3^{rd}$ root is -5 so we have the product of 2 roots = 5 (as product of all roots -25)

$x^3-xk+25 = (x+5)(x^2- 5x + 5) = x^3 -5x^2+5x + 5x^2 - 25 x + 25 = x^3 -20x + 25$

comparing with given equation we have k = 20 and hence $| k | = 20$

we have to check that $x^2-5x+5=0$ is having real roots or not. We have discriminant = $5^2-20= 5 >0 $ so real roots

so ans 20

 

[topsquark]

[sp]
There's a typo.

\(\displaystyle x^3 - k x + 25 = x^3 - 20 x + 25 \implies k = 20\), not -20.
[/sp]
-Dan

 

[kaliprasad]

[sp]
There's a typo.

\(\displaystyle x^3 - k x + 25 = x^3 - 20 x + 25 \implies k = 20\), not -20.
[/sp]
-Dan

Thanks I have done the needful in line for the flow