Question 1209706: There are integers b, c for which both roots of the polynomial x^2 - x - 3 are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Let $P(x) = x^2 - x - 3$. The roots of $P(x)$ are given by the quadratic formula:
$$x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-3)}}{2(1)} = \frac{1 \pm \sqrt{1+12}}{2} = \frac{1 \pm \sqrt{13}}{2}$$
Let $r_1 = \frac{1 + \sqrt{13}}{2}$ and $r_2 = \frac{1 - \sqrt{13}}{2}$.
Since $r_1$ and $r_2$ are roots of $x^2 - x - 3 = 0$, we have
$r_1^2 - r_1 - 3 = 0$ and $r_2^2 - r_2 - 3 = 0$.
Thus, $r_1^2 = r_1 + 3$ and $r_2^2 = r_2 + 3$.
Also, $r_1 + r_2 = 1$ and $r_1 r_2 = -3$.
Let $Q(x) = x^3 - bx^2 - c$. Since $r_1$ and $r_2$ are roots of $Q(x)$, we have
$r_1^3 - br_1^2 - c = 0$ and $r_2^3 - br_2^2 - c = 0$.
$r_1^3 = r_1 \cdot r_1^2 = r_1(r_1+3) = r_1^2 + 3r_1 = (r_1+3) + 3r_1 = 4r_1 + 3$
$r_2^3 = r_2 \cdot r_2^2 = r_2(r_2+3) = r_2^2 + 3r_2 = (r_2+3) + 3r_2 = 4r_2 + 3$
Substituting these into the equations for $Q(x)$, we get
$4r_1 + 3 - b(r_1 + 3) - c = 0$
$4r_2 + 3 - b(r_2 + 3) - c = 0$
Subtracting the second equation from the first, we get
$4(r_1 - r_2) - b(r_1 - r_2) = 0$
$r_1 - r_2 \ne 0$, so $4 - b = 0$, which gives $b = 4$.
Substituting $b=4$ into the first equation, we get
$4r_1 + 3 - 4(r_1 + 3) - c = 0$
$4r_1 + 3 - 4r_1 - 12 - c = 0$
$-9 - c = 0$
$c = -9$
Thus, $(b, c) = (4, -9)$.
Final Answer: The final answer is $\boxed{(4,-9)}$
Answer by ikleyn(52925)  (Show Source):
You can put this solution on YOUR website! .
There are integers b, c for which both roots of the polynomial x^2 - x - 3
are also roots of the polynomial x^3 - bx^2 - c. Determine the ordered pair (b,c).
~~~~~~~~~~~~~~~~~~~~~~~~~
Let p and q be the roots of the polynomial x^2 - x - 3.
Due to Vieta's theorem,
p + q = 1, (1)
pq = -3. (2)
According to the problem, p and q are also the roots of the polynomial x^3 - bx^2 - c.
Let r be the third root of this polynomial.
Then, due to Vieta's theorem for polynomial x^3 - bx^2 - c
p + q + r = b, (3)
p*q + p*r + q*r = 0, (4) (coefficient at x in polynomial x^3 - bx^2 - c)
p*q*r = c. (5)
In (3), replace p+q by 1, based on (3). In (5), replace p*q by -3, based on (2).
Then from (3) an (4) you will have
1 + r = b, (6)
-3r = c, (7)
In equation (4), replace p*q by -3, based on (2). Then equation (4) takes the form
-3 + pr + qr = 0,
or
p*r + q*r = 3,
(p + q)*r = 3.
In the last equation, replace (p + q) by 1, based on (1). You will get
1*r = 3, i.e. r = 3.
Now from (6) b = 1 + r = 1 + 3 = 4;
from (7) c = -3r = -3*3 = -9.
ANSWER. In polynomial x^3 - bx^2 - c, coefficients are b = 4, c = -9.
At this point, the problem is solved completely.
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