# Viet Theorem

## Related calculator: Quadratic Equation Calculator

**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 $$${4}$$$.