Rational root coefficient

Question

I saw this question in my exam recently,

If a, b, c are distinct rational roots of $x^3+ax^2+bx+c=0$, find the values of a, b, c.

Can someone give me a hint or answer? I tried factoring it and then expand it all out, but it doesn't seem very useful as they are all "rational". Can someone help me with it?

Answer 1

Apologies for the length of the argument. I tried to format it to be more comprehensible.

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Multiplying out $(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+ac+bc)x-abc$, so we have the equations:

$1)$ $a=-a-b-c$

$2)$ $b=ab+ac+bc$

$3)$ $c=-abc$

Notice that if $a = 0$ or $b = 0$, then by the $(3)$, $c=0$ contradicting the fact that they must be distinct. Therefore $a,b\neq 0$.

Now if $c=0$, then by $(1)$ and $(2)$, $2a=-b$ and $b=ab$, which has the solution $a=1, b=-2, c=0$. The goal now is to show no other solution exists.

Consider when $c \neq 0$. Cancelling the $c$'s in $(3)$ gives us $1=-ab \implies a=-\frac{1}{b}$. Then we can simplify $(1)$ into $b^2+cb-2=0$

$\implies b=\frac{-c \pm \sqrt{c^2+8}}{2}$

and $(2)$ simplifies to $(c-1)b^2-b-c=0$

$\implies b=\frac{1\pm\sqrt{4c^2-4c+1}}{2(c-1)}=\frac{1\pm(2c-1)}{2(c-1)}$

for $c\neq 1$

(if $c=1$, then $(c-1)b^2-b-c=0\implies b=-1 \implies a=1$, contradicting distinct values again).

Setting the equations for $b$ equal yields

$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\pm(2c-1)}{2(c-1)}$

or

$\frac{-c \pm \sqrt{c^2+8}}{2}=\frac{1\mp(2c-1)}{2(c-1)}$

Changing these two equations to the one

$c^2+8=\left(\frac{1\pm(2c-1)}{c-1}+c \right)^2$

makes it slightly easier.

Taking the $-$ gives us $c=-1 \implies b=-1$ contradicting distinct again, and taking the $+$ gives us the more ugly cubic equation

$c^3-2c^2+4c-2=0$

By the Rational Root Theorem, one need only test the values $\pm 1, \pm 2$, none of which yield a solution to this cubic, so this cubic has no rational solutions.

Therefore, the only solution was the first one we found; namely $\boxed{a=1, b=-2,c=0}$.

Answer 2

Try sum of roots, product of roots of cubic equation

$\alpha + \beta + \gamma = -\frac ba $

$\alpha\beta + \beta\gamma + \alpha\gamma = \frac ca $

$\alpha \beta \gamma = -\frac da $

Answer 3

The idea provided by @JasonM is correct. You should be able to solve the problem as follows.

Notice that the cubic polynomial is monic, meaning that the coefficient of $x^3$ is $1.$ This is nice because now we can write the polynomial as $(x - a)(x - b)(x - c)$ in factored form. Now, we can equate this to the given form: $$(x - a)(x - b)(x - c) = x^3 + ax^2 + bx + c$$ $$x^3 - (a + b + c)x^2 + (ab + bc + ca)x - abc = x^3 + ax^2 + bx + c.$$

We can see from the equation that $-abc = c,$ meaning that $ab = -1.$ We also have that $a + b + c = a,$ meaning that $b + c = 0,$ or $b = -c.$ With $ab + bc + ca = b,$ substituting values in, we get $-1 - c^2 + 1 = - c,$ which can be rearranged to $c^2 - c = c(c - 1) = 0.$ The solution $c = 0$ cannot be accepted because in that case, $b = c = 0$ and the roots are not distinct. But if $c = 1,$ then $b = -1$ and $a = 1,$ in which case $a = c.$ So there is $\boxed{\text{No solution}}$ to the requested problem.