Let
$$p(x) = ax^2 + bx + c, \quad a \neq 0$$
be the second degree polynomial. We can factorize it in the following way:
$$p(x) = a (x – x_1 ) (x-x_2),$$
where $x_1$ and $x_2$ are roots of the second degree polynomial.
By equating two polynomials above we obtain:
$$ax^2 + bx + c = a ( x^2 – x_2 x – x_1 x + x_1 x_2)$$
$$ax^2 + bx + c = a x^2 – a x_2x – a x_1 x + a x_1 x_2$$
$$ax^2 + bx + c = ax^2 – a(x_1 + x_2)x + a x_1 x_2$$
By the theorem of equality of polynomials (two polynomials are equal if their coefficients by the proper powers are equal) we obtain:
$$b = a (x_1 + x_2) \Rightarrow x_1 + x_2 = \displaystyle{- \frac{b}{a}}$$
and
$$ c = a \cdot x_1 x_2 \Rightarrow x_1 x_2 = \displaystyle{\frac{c}{a}}.$$
Let
$$p(x) = a x^3 + b x^2 + c x + d, \quad a \neq 0$$
be the third degree polynomial. We can factorize it in the following way:
$$p(x) = a (x – x_1) (x-x_2) (x – x_3),$$
where $x_1, x_2$ and $x_3$ are its roots.
Analogous to the procedure described for a second degree polynomial, we obtain:
$$x_1 + x_2 + x_3 = \displaystyle{- \frac{b}{a}},$$
$$x_1x_2 + x_1 x_3 + x_2 x_3 = \displaystyle{\frac{c}{a}},$$
$$x_1 x_2 x_3 = \displaystyle{- \frac{d}{a}}.$$
Example 1. For the equation
$$x^2 + 7x -1 = 0$$
determine $(x_1 + x_2)^2$ and $\displaystyle{\frac{1}{x_1} + \frac{1}{x_2}}$, where $x_1$ and $x_2$ are roots of the given equation.
Solution :
By using the Vieta’s formulas, we have:
$$x_1 + x_2 = -7,$$
$$x_1x_2 = 1.$$
Now we have:
$$(x_1 + x_2)^2 = 7 ^2 = 49.$$
On the other side
$$\frac{1}{x_1} + \frac{1}{x_2} = \frac{x_2 + x_1}{x_1x_2} = \frac{7}{1} =7.$$
The application of the Vieta’s formulas for solving the system of equations
Example 2. Solve the following system of equations:
$$x + y + z = 2$$
$$x^2 + y^2 + z^2 = 6$$
$$x^3 + y^3 + z^3 = 8.$$
Solution :
Since
$$(x +y + z) ^2 = x^2 + y^2 + z^2 + 2 (xy + yz + xz)$$
$$\Rightarrow x^2 + y^2 + z^2 = (x + y + z)^2 – 2 (xy + yz + xz) $$
then, because $(x + y + z)^2 = 4$ and $x^2 + y^2 + z^2 = 6$:
$$ 6 = 4 – 2 (xy + yz + xz) \Rightarrow 2 = – 2 (xy + yz + xz)$$
$$\Rightarrow xy + yz + xz = -1.$$
Since
$$(x + y + z)^3 =(x^3 + y^3 + z ^3) + 3 ( x + y + z) ( xy + xz + yz) -3 xyz$$
it follows
$$ 2 ^3 = 8 + 3 \cdot 2 \cdot -1 – 3xyz,$$
that is,
$$6 = – 3xyz \Rightarrow xyz = -2.$$
The given system of equations is equivalent to the following system:
$$ x + y + z = 2$$
$$ xy + yz + xz = -1$$
$$ xyz = -2.$$
By the Vieta’s formulas now we have:
$$ – a = 2 $$
$$ b = -1$$
$$-c = -2.$$
Substituting $a, b$ and $c$ in the third degree polynomial $p(t) = t^3 + a t ^2 + bt +c$, we obtain:
$$p(t) = t^3 – 2 t^2 – t +2.$$
If $t = \displaystyle{\frac{p}{q}}$, $q \neq 0$, is a rational root of $p(t)$, then $p$ is a factor of the free coefficient and $q$ is a factor of the leading coefficient.
If this polynomial has rational roots, then $ t \in \{ -2 , -1 , 1 , 2\}$.
For $t = 1$ we have:
$$p(1) = 1^3 – 2 \cdot 1^2 – 1 + 2 = 1 – 2 -1 + 2 = 0,$$
that is $t=1$ is the one root of the polynomial $p(t) = t^3 – 2 t^2 – t +6.$ The rest two we obtain by dividing $p(t)$ with $f(t)=t-1$:
$$t^3 – 2 t^2 – t +2 : t – 1 = t^2 – t -2,$$
that is, roots of the $h(t) = t^2 – t -2$ are:
$$t_{2,3} = \displaystyle{\frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3 }{2}}$$
$$\Rightarrow t_2 = -1 , t_3 = 2.$$
Finally,
$$p(t) = f(t)\cdot h(t)$$
$$p(t) = (t-2) (t-1) (t+1).$$
The system is symmetrical in relation to the $x, y, z$ , so we have $6$ solutions:
$$(x, y, z) \in \{( -1, 1, 2) , (-1, 2, 1) , (2, -1, 1) , (2, 1, -1), ( 1, -1, 2), (1, 2, -1)\}.$$
