Theory

Polynomial and rational equations and inequalities

How to solve first- and second-degree equations, how to lower the degree by factorization, and how to study inequalities with the sign of the parabola, the sign chart and systems.

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First-degree equations
Second-degree equations

First-degree equations

First-degree equation

The equation $a x + b = 0$ with $a \neq = 0$ has the unique solution

x = - \frac{b}{a} .

This form is reached with the equivalence principles: adding the same quantity to both sides and multiplying both sides by the same nonzero quantity produce equations with the same solutions. For $a = 0$ the equation reduces to $b = 0$ and there are two cases:

$b \neq = 0$ : no solution (impossible equation);
$b = 0$ : every $x$ is a solution (identity).

For example, $3 (x - 2) + 5 = 2 x$ is equivalent to $3 x - 1 = 2 x$ , hence $x = 1$ .

Second-degree equations

1. Quadratic formula

The equation $a x^{2} + b x + c = 0$ with $a \neq = 0$ has discriminant $Δ = b^{2} - 4 a c$ . If $Δ \geq 0$ the solutions are

x = \frac{- b \pm Δ}{2 a};

if $Δ < 0$ there are no real solutions. The formula solves every second-degree equation, therefore every polynomial equation is considered solved once it is reduced to factors of degree at most two.

Proof

Since $a \neq = 0$ , multiplying both sides by $4 a$ gives the equivalent equation $4 a^{2} x^{2} + 4 ab x + 4 a c = 0$ . The first two terms can be completed to a square, since $4 a^{2} x^{2} + 4 ab x = (2 a x + b)^{2} - b^{2}$ , so the equation can be rewritten as

(2 a x + b)^{2} = b^{2} - 4 a c = Δ.

If $Δ < 0$ no $x$ satisfies it, because a square is never negative. If $Δ \geq 0$ , the equality holds when $2 a x + b = \pm Δ$ , therefore

x = \frac{- b \pm Δ}{2 a} .

2. Discriminant

The sign of $Δ = b^{2} - 4 a c$ counts the real solutions:

$Δ > 0$ : two distinct real solutions $x_{1} \neq = x_{2}$ ;
$Δ = 0$ : one double real solution $x = - \frac{b}{2 a}$ ;
$Δ < 0$ : no real solution.

For $x^{2} - 2 x - 3 = 0$ we have $Δ = 4 + 12 = 16 > 0$ and the solutions are $x = \frac{2 \pm 4}{2}$ , that is, $3$ and $- 1$ . For $x^{2} + 2 x + 1 = 0$ we have $Δ = 0$ and $x = - 1$ double. For $x^{2} + x + 1 = 0$ we have $Δ = - 3 < 0$ and no real solution.

3. Incomplete equations

The quadratic formula also applies to equations with $b = 0$ or $c = 0$ ; these cases are solved directly with fewer computations.

Equation without constant term $a x^{2} + b x = 0$ : factoring out $x$ gives $x (a x + b) = 0$ , with solutions $x = 0$ and $x = - \frac{b}{a}$ .
Equation without linear term $a x^{2} + c = 0$ : it is equivalent to $x^{2} = - \frac{c}{a}$ ; if $- \frac{c}{a} > 0$ the solutions are $x = \pm - \frac{c}{a}$ , if $- \frac{c}{a} = 0$ the only solution is $x = 0$ , if $- \frac{c}{a} < 0$ there are no real solutions.

For example, $3 x^{2} - 12 = 0$ is equivalent to $x^{2} = 4$ , so $x = \pm 2$ .

4. Sum and product of the roots

If $Δ \geq 0$ , the roots $x_{1}$ and $x_{2}$ of $a x^{2} + b x + c = 0$ (possibly coincident) satisfy

x_{1} + x_{2} = - \frac{b}{a}, x_{1} \cdot x_{2} = \frac{c}{a},

and the trinomial factors as

a x^{2} + b x + c = a (x - x_{1}) (x - x_{2}) .

The two equalities follow from the quadratic formula, by adding and multiplying the expressions of the roots. With them one finds integer roots mentally and factors trinomials: for $x^{2} - 7 x + 12$ two numbers with sum $7$ and product $12$ are $3$ and $4$ , so the roots are $3$ and $4$ and the trinomial is $(x - 3) (x - 4)$ .

The theory continues with 6 more sections (higher degree equations, first degree inequalities, second degree inequalities…), plus all worked examples, the solutions and the PDF.

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