Learn/Algebra/Solving Algebraic Equations

Algebra • Topic 2

Solving Algebraic Equations

Beyond the quadratic formula, olympiad equations often require substitution, symmetry, or identifying hidden structures.

Quadratics and Reducibles

Theorem (Quadratic Formula)

The roots of

a x^{2} + b x + c = 0

are:

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

Many higher-degree equations are "quadratics in disguise."

Bi-quadratic: $a x^{4} + b x^{2} + c = 0$ . Let $u = x^{2}$ .
Reciprocal: $a x^{4} + b x^{3} + c x^{2} + b x + a = 0$ . Divide by $x^{2}$ and let $u = x + \frac{1}{x}$ .

For systems involving sums and products, changing variables to the elementary symmetric polynomials often simplifies the algebra. Let:

e_{1} = x + y, e_{2} = x y

Then

x^{2} + y^{2} = e_{1}^{2} - 2 e_{2}

, and

x^{3} + y^{3} = e_{1}^{3} - 3 e_{1} e_{2}

Example

Solve the system:

Proof (Solution). Using $e_{1} = 5$ , the second equation becomes:

e_{1}^{3} - 3 e_{1} e_{2} = 35 ⟹ 125 - 15 e_{2} = 35

15 e_{2} = 90 ⟹ e_{2} = 6

x + y = 5

and

x y = 6

. The solutions are roots of

t^{2} - 5 t + 6 = 0

, which are

t = 2, 3

. Solution pairs:

(2, 3)

and

(3, 2)

. ∎

Warning (Extraneous Solutions)

When squaring both sides of an equation to remove radicals, you may introduce false solutions. Always check your answers in the original equation.

If you see terms like

A - B = C

, multiplying by the conjugate

A + B

can reveal a simpler system, since

(A - B) (A + B) = A - B

Exercise (Problem 1)

Solve for real

x

x + 3 - 4 x - 1 + x + 8 - 6 x - 1 = 1

Exercise (Problem 2)

Solve the system for real numbers:

Exercise (Problem 3)

Solve for

x

(x - 1) (x - 2) (x - 3) (x - 4) = 24

. (Hint: Group $(x - 1) (x - 4)$ and $(x - 2) (x - 3)$ )