# Viet Theorem

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Viet Theorem. If quadratic equation ax^2+bx+c=0 (reduced form is x^2+b/a+c/a=0) has roots p and q, then color(green)(p+q=-b/a), color(magenta)(pq=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 (x-p)(x-q)=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 (x-p)(x-q)=0 has two roots: p and q.

Next, apply FOIL to the left hand side:

(x-p)(x-q)=x^2-px-qx+pq=x^2-(p+q)x+pq.

We've got equivalent equation x^2-(p+q)x+px=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+b/ax+c/a=0, when -(p+q)=b/a and pq=c/a.

From the last two equalities follows Viet Theorem.

Example 1. Without solving equation ax^2+bx+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=-b/a and pq=c/a.

We need to complete the square: p^2+q^2=p^2+q^2color(red)(+2pq-2pq)=(p+q)^2-2pq=(-b/a)^2-2c/a=b^2/a^2-(2ac)/a^2=(b^2-2ac)/a^2.

Answer: (b^2-2ac)/(a^2).

Inverse Viet Theorem. If numbers p and q are such, that p+q=-b/a and pq=c/a, then p and q are roots of the equation ax^2+bx+c=0.

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

Let's try to find such two numbers p and q that p+q=-3 and pq=-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.