Spurious Quadratic Equations
A spurious quadratic equation is a second-degree equation of the form ax2 + bx + c = 0 where the constant term c = 0, while the coefficients a and b are both nonzero. $$ ax^2 + bx = 0 $$
How to solve a spurious quadratic equation
To solve this type of equation, the first step is to factor out the variable x:
$$ x \cdot (ax + b) = 0 $$
We then apply the zero-product property.
The equation x(ax + b) = 0 holds true either when x = 0 (the trivial solution) or when ax + b = 0.
$$ x = 0 $$
$$ x = - \frac{b}{a} $$
Therefore, a spurious quadratic equation always has two real solutions, and one of them is always zero (x = 0).
Note. This is the fastest and most straightforward method for finding the real solutions of a spurious quadratic equation.
A worked example
Let's consider the following quadratic equation:
$$ 2x^2 + 4x = 0 $$
We start by factoring out x from the left-hand side:
$$ x \cdot (2x + 4) = 0 $$
Now, applying the zero-product property gives us two possible cases.
The first real solution is x = 0:
$$ x = 0 $$
Note. When x = 0, the product 0·(2x + 4) = 0 is always equal to zero. This solution is called the trivial solution of the spurious equation.
The second real solution is obtained by solving the linear equation corresponding to the second factor:
$$ 2x + 4 = 0 $$
$$ x = - \frac{4}{2} = -2 $$
Thus, the second solution is x = -2.
Note. When x = -2, the term (2x + 4) becomes zero, and the product x(2x + 4) = 0 reduces to x·0 = 0 because the second factor vanishes. The product of any number and zero is always zero.
In conclusion, the two real solutions of this spurious quadratic equation are x = 0 and x = -2.
Proof
Let's start from the general form of a spurious quadratic equation:
$$ ax^2 + bx = 0 $$
We can rewrite it in an equivalent form by factoring out x:
$$ x \cdot (ax + b) = 0 $$
In this form, the equation is expressed as the product of two factors: x and (ax + b).
The first (trivial) solution
If x = 0, the first factor is zero, and any product involving zero is always zero:
$$ 0 \cdot (ax + b) = 0 $$
Hence, the equation ax2 + bx = 0 is satisfied.
The second solution
If ax + b = 0, the second factor is zero, and once again any product involving zero equals zero:
$$ x \cdot \underbrace{(ax + b)}_{0} = 0 $$
$$ x \cdot 0 = 0 $$
So the equation ax2 + bx = 0 is satisfied in this case as well.
To find the value of x that makes the second factor zero, we solve the linear equation ax + b = 0:
$$ ax + b = 0 $$
$$ x = - \frac{b}{a} $$
Therefore, x = -b/a is the second real solution of the spurious quadratic equation ax2 + bx = 0.
And so on.
