When you have an equation of the form \( ax^2 + bx + c = d \), and you set one side to zero, it becomes \( ax^2 + bx + (c - d) = 0 \).
If we consider the case where the numeric expression on the right side is zero (i.e., \( c - d = 0 \)), the equation simplifies to \( ax^2 + bx = 0 \).
Factoring this yields:
\[ x(ax + b) = 0 \]
This gives us two conditions to solve:
- \( x = 0 \)
- \( ax + b = 0 \) (which gives a second solution \( x = -\frac{b}{a} \), provided \( a \neq 0 \))
From this analysis, if \( a \neq 0 \), there will generally be two solutions (including \( x = 0 \) and \( x = -\frac{b}{a} \)). However, if the equation is a perfect square (for instance, if \( b = 0 \)), both solutions could be the same, resulting in one unique solution.
In conclusion:
- If \( a = 0 \), you would get either no solution or infinite solutions, depending on the other terms.
- If \( a \neq 0 \), then when the expression equals zero, there are either one or two solutions, depending on whether the factors produce a double root or two distinct roots.
Based on the statements:
- Statement 1: If the numeric expression is zero, there are two solutions. (Incorrect for the general case).
- Statement 2: If the numeric expression is zero, there is one solution. (Might also be correct in the case of a perfect square).
- Statement 3: If the numeric expression is zero, there are no solutions. (Incorrect for this scenario).
Thus, the correct perspective is that it can lead to one or two solutions depending on the specifics of the coefficients involved, but typically, for a standard quadratic \( ax^2 + bx = 0 \) (assuming \( a \neq 0 \)), you will find two solutions.
Therefore, one might argue that none of the statements are wholly correct without providing additional details on \( a \) and \( b \). If we had to choose, Statement 2 could be the least inaccurate in cases where a double root is present but does not encompass all scenarios accurately.
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