Incomplete Quadratic Equations

Definition

1.A quadratic equation is incomplete when one or both of the terms $bx$ and $c$ are absent from the standard form $ax^2 + bx + c = 0$, provided the term $ax^2$ is present. These equations admit direct solution methods that do not require the quadratic formula or factorization.

2.When both $b$ and $c$ are equal to zero, the equation reduces to:

$$ax^2 = 0, \quad a \neq 0$$

3.Dividing both sides by $a$, which is nonzero by assumption, gives $x^2 = 0$, and the only real solution is $x = 0$ for every admissible value of $a$.

4.

Graphically, the equation represents a parabola with vertex at the origin $(0, 0)$, symmetric about the $y$-axis. The graph opens upward when $a > 0$ and downward when $a < 0$, and the magnitude of $a$ determines its width. Although the equation has a single solution, the function has a double root at $x = 0$: the $x$-axis is tangent to the parabola at the origin.

The case b = 0

5.When $b = 0$, the equation takes the form:

$$ax^2 + c = 0, \quad a \neq 0, \ c \neq 0$$

6.Isolating $x^2$ gives:

$$x^2 = -\frac{c}{a}$$

7.The equation represents a parabola symmetric about the $y$-axis. When $a$ and $c$ have opposite signs, the quantity $-\frac{c}{a}$ is positive and the equation has two distinct real solutions:

$$x_{1,2} = \pm\sqrt{-\frac{c}{a}}$$

8.The parabola then intersects the $x$-axis at two points symmetric with respect to the origin. When $a$ and $c$ have the same sign, the quantity $-\frac{c}{a}$ is negative and the equation has no real solution:

$$-\frac{c}{a} < 0 \implies x \notin \mathbb{R}$$

9.In this case the parabola lies entirely above or below the $x$-axis and does not intersect it.

The case c = 0

10.When $c = 0$, the equation takes the form:

$$ax^2 + bx = 0, \quad a \neq 0, \ b \neq 0$$

11.Factoring out $x$ gives $x(ax + b) = 0$. By the zero product property, either $x = 0$ or $ax + b = 0$, so the equation has two distinct real solutions:

$$x_1 = 0 \qquad x_2 = -\frac{b}{a}$$

Examples

12.Consider the equation $3x^2 = 0$. Since both $b$ and $c$ are zero, the only solution is $x = 0$.

13.Consider the equation $2x^2 - 8 = 0$. This has the form $ax^2 + c = 0$ with $a = 2$ and $c = -8$. Since $a$ and $c$ have opposite signs, two real solutions exist. Isolating $x^2$ gives:

$$x^2 = \frac{8}{2} = 4$$

14.Taking the square root of both sides yields the two solutions:

$$x_{1,2} = \pm\sqrt{4} = \pm 2$$

15.Consider the equation $x^2 + 5 = 0$. Here $a = 1$ and $c = 5$ have the same sign, so $-\frac{c}{a} = -5 < 0$. The equation has no real solutions.

16.Consider the equation $3x^2 - 6x = 0$. This has the form $ax^2 + bx = 0$ with $a = 3$ and $b = -6$. Factoring out $x$ gives $x(3x - 6) = 0$, and the two solutions are:

$$x_1 = 0 \qquad x_2 = \frac{6}{3} = 2$$

A common error to avoid

17.For equations of the form $ax^2 + bx = 0$, a frequent error consists in dividing both sides by $x$ when the equation is written as:

$$ax^2 = bx$$

18.Dividing by $x$ is not valid here, since $x = 0$ is itself a solution and division by zero is undefined. This manipulation eliminates the root $x = 0$ and reduces the equation to a linear one, producing only the solution $x = -\frac{b}{a}$. The correct approach is to collect $x$ as a common factor, as shown above.