Viet Theorem

Viet Theorem. If quadratic equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$ (reduced form is ${{x}}^{{2}}+\frac{{b}}{{a}}+\frac{{c}}{{a}}={0}$) has roots ${p}$ and ${q}$, then ${\color{green}{{{p}+{q}=-\frac{{b}}{{a}}}}}$, ${\color{ma\genta}{{{p}{q}=\frac{{c}}{{a}}}}}$, i.e. sum of the roots equals second coefficient, taken with opposite sign, and product of roots equals constant.

Indeed, let's start from equation ${\left({x}-{p}\right)}{\left({x}-{q}\right)}={0}$. Note, that ${p}$ and ${q}$ are some numbers.

Product equals 0, only when at least one factor equals 0.

In other words, either ${x}-{p}={0}$ and ${x}-{q}={0}$.

Thus, equation ${\left({x}-{p}\right)}{\left({x}-{q}\right)}={0}$ has two roots: ${p}$ and ${q}$.

Next, apply FOIL to the left hand side:

${\left({x}-{p}\right)}{\left({x}-{q}\right)}={{x}}^{{2}}-{p}{x}-{q}{x}+{p}{q}={{x}}^{{2}}-{\left({p}+{q}\right)}{x}+{p}{q}$.

We've got equivalent equation ${{x}}^{{2}}-{\left({p}+{q}\right)}{x}+{p}{x}={0}$.

Original equation will have roots ${p}$ and ${q}$, when coefficients of above equation equal coefficients of the original equation.

In other words, ${p}$ and ${q}$ are roots of the equation ${{x}}^{{2}}+\frac{{b}}{{a}}{x}+\frac{{c}}{{a}}={0}$, when $-{\left({p}+{q}\right)}=\frac{{b}}{{a}}$ and ${p}{q}=\frac{{c}}{{a}}$.

From the last two equalities follows Viet Theorem.

Example 1. Without solving equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$, find ${{p}}^{{2}}+{{q}}^{{2}}$, in terms of ${a}$, ${b}$ and ${c}$, where ${p}$ and ${q}$ are unknown roots of the equation.

By the Viet Theorem: ${p}+{q}=-\frac{{b}}{{a}}$ and ${p}{q}=\frac{{c}}{{a}}$.

We need to complete the square: ${{p}}^{{2}}+{{q}}^{{2}}={{p}}^{{2}}+{{q}}^{{2}}{\color{red}{{+{2}{p}{q}-{2}{p}{q}}}}={{\left({p}+{q}\right)}}^{{2}}-{2}{p}{q}={{\left(-\frac{{b}}{{a}}\right)}}^{{2}}-{2}\frac{{c}}{{a}}=\frac{{{b}}^{{2}}}{{{a}}^{{2}}}-\frac{{{2}{a}{c}}}{{{a}}^{{2}}}=\frac{{{{b}}^{{2}}-{2}{a}{c}}}{{{a}}^{{2}}}$.

Answer: $\frac{{{{b}}^{{2}}-{2}{a}{c}}}{{{{a}}^{{2}}}}$.

Inverse Viet Theorem. If numbers ${p}$ and ${q}$ are such, that ${p}+{q}=-\frac{{b}}{{a}}$ and ${p}{q}=\frac{{c}}{{a}}$, then ${p}$ and ${q}$ are roots of the equation ${a}{{x}}^{{2}}+{b}{x}+{c}={0}$.

Example 2. Solve the equation ${{x}}^{{2}}+{3}{x}-{28}={0}$.

Let's try to find such two numbers ${p}$ and ${q}$ that ${p}+{q}=-{3}$ and ${p}{q}=-{28}$.

It is not hard to see, that such numbers are $-{7}$ and ${4}$. Thus, by the Inverse Viet Theorem, they are the roots of the given equation.

Answer: $-{7}$ and ${4}$.