Bi-Quadratic Equations

A bi-quadratic equation is a fourth-degree polynomial equation that contains only even powers of the variable.

Definition

The general form is:

$ax^4 + bx^2 + c = 0$

where:
$(a \neq 0)$ $a, b, c$ are real numbers

There are no $x^3$ or $x$ terms. Because only even powers appear, the equation can be reduced to a quadratic equation.

Main Idea: Substitution

Let:

$u = x^2$

Then the equation becomes:

$au^2 + bu + c = 0$

This is a standard quadratic equation.

After solving it, we return to:
$x^2 = u$

Since we are working in the real numbers, we must remember:
$x^2 \ge 0$
Negative values of $u$ do not produce real solutions.

General Solution Procedure

Step 1: Substitute $u = x^2$

Step 2: Solve the quadratic equation in $u$

Step 3: Keep only solutions where $u \ge 0$

Step 4: Solve
$x = \pm \sqrt{u}$

Example 1 — Four Real Solutions

Solve:
$x^4 - 5x^2 + 4 = 0$

Substitute: $u^2 - 5u + 4 = 0$

Factor:
$(u - 4)(u - 1) = 0$
$u = 4 \quad \text{or} \quad u = 1$

Both are nonnegative.

Back-substitute:
$x^2 = 4 \Rightarrow x = \pm 2$

$x^2 = 1 \Rightarrow x = \pm 1$

Solutions:
$x = -2, -1, 1, 2$

Example 2 — Two Real Solutions (One Positive, One Negative Value of $u$)

Solve:
$x^4 - x^2 - 6 = 0$

Substitute: $u^2 - u - 6 = 0$

Factor: $ (u - 3)(u + 2) = 0$

$u = 3 \quad \text{or} \quad u = -2$

Since $x^2 \ge 0$, we reject $u = -2$.

Solve the valid case: $x^2 = 3$
$x = \pm \sqrt{3}$

Solutions: $x = -\sqrt{3}, \quad \sqrt{3}$

Example 3 — One Real Solution (Repeated Root)

Solve: $x^4 - 4x^2 = 0$

Factor: $x^2(x^2 - 4) = 0$
$x^2 = 0 \quad \text{or} \quad x^2 = 4$

Solve:
$x = 0$
$x = \pm 2$

Solutions:
$x = -2, 0, 2$

Here, $x = 0$ is a repeated root.
All roots are: $-2, 0, 0, 2$

Example 4 — No Real Solutions

Solve: $x^4 + 4x^2 + 5 = 0$

Substitute: $u^2 + 4u + 5 = 0$

Discriminant: $\Delta = 16 - 20 = -4$

Since the quadratic has no real solutions, the original equation has no real solutions.

Summary of All Possible Cases

Let the quadratic in $u$ have solutions $u_1$ and $u_2$.

Graphical Interpretation

The function $f(x) = ax^4 + bx^2 + c$

is symmetric about the y-axis because:
$f(x) = f(-x)$

The real solutions correspond to the points where the graph intersects the x-axis.