# Incomplete Quadratic Equations
Source: algebrica.org — CC BY-NC 4.0
https://algebrica.org/quadratic-equations/
## Definition
A [quadratic equation](../quadratic-equations.md) is considered 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](../quadratic-formula.md) or [factorization](../factoring-quadratic-equations.md).
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When both \\(b\\) and \\(c\\) are equal to zero, the equation reduces to:
\\[ax^2 = 0, \quad a \neq 0\\]
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\\).
> Graphically, the equation represents a [parabola](../parabola.md) with its vertex at the origin \\((0, 0)\\), symmetric about the y-axis. The graph opens upward if \\(a > 0\\) and downward if \\(a < 0\\); the magnitude of \\(a\\) determines the width of the parabola. 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.
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## The case b = 0
When \\(b = 0\\), the equation takes the form:
\\[ax^2 + c = 0, \quad a \neq 0,\\, c \neq 0\\]
Isolating \\(x^2\\) gives:
\\[x^2 = -\frac{c}{a}\\]
The equation represents a parabola symmetric about the y-axis. When \\(a\\) and \\(c\\) have opposite signs, the quantity \\(-c/a\\) is positive and the equation has two distinct real solutions:
\\[x_{1,2} = \pm\sqrt{-\frac{c}{a}}\\]
The parabola intersects the x-axis in two points symmetric with respect to the origin. When \\(a\\) and \\(c\\) have the same sign, the quantity \\(-c/a\\) is negative and the equation has no real solutions:
\\[-\frac{c}{a} < 0 \implies x \notin \mathbb{R}\\]
The parabola lies entirely above or below the x-axis and does not intersect it.
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## The case c = 0
When \\(c = 0\\), the equation takes the form:
\\[ax^2 + bx = 0, \quad a \neq 0,\\, b \neq 0\\]
Factoring out \\(x\\) gives \\(x(ax + b) = 0\\). Applying the zero product property, either \\(x = 0\\) or \\(ax + b = 0\\), and the equation has two distinct real solutions:
\\[x_1 = 0 \qquad x_2 = -\frac{b}{a}\\]
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## Examples
Consider the equation \\(3x^2 = 0\\). Since both \\(b\\) and \\(c\\) are zero, the only solution is \\(x = 0\\).
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Consider the equation \\(2x^2 - 8 = 0\\). This is of 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\\]
Taking the square root of both sides yields the two solutions:
\\[x_{1,2} = \pm\sqrt{4} = \pm 2\\]
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Consider the equation \\(x^2 + 5 = 0\\). Here \\(a = 1\\) and \\(c = 5\\) have the same sign, so \\(-c/a = -5 < 0\\). The equation has [no real solutions](../quadratic-equations-with-complex-solutions.md).
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Consider the equation \\(3x^2 - 6x = 0\\). This is of 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\\]
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## A common error to avoid
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\\]
Dividing by \\(x\\) is not a valid operation here, since \\(x = 0\\) is itself a solution and division by zero is undefined. This manipulation [eliminates the root \\(x = 0\\)](../loss-of-roots.md) and reduces the equation to a linear one, producing only the solution \\(x = -b/a\\). The correct approach is to collect \\(x\\) as a common factor, as shown above.
