To find which equations have the solution set of {12, -3}, we can use the fact that if a polynomial equation has a certain set of roots, it can be factored accordingly. The roots can be expressed in the form of factors:
If the roots are \( x = 12 \) and \( x = -3 \), then the corresponding factors will be \( (x - 12) \) and \( (x + 3) \).
Thus, the polynomial can be represented as:
\[ (x - 12)(x + 3) = 0 \]
Let's analyze the given options:
-
(4x - 2)(x + 3) = 0:
- This does not yield a root of 12 or -3.
-
(x - 12)(x + 3) = 0:
- This matches perfectly with our expected factors, yielding the roots 12 and -3.
-
(3x - 6)(3x - 6) = 0:
- This simplifies to \( (3(x - 2))^2 = 0 \), giving x = 2 as the only solution.
-
(6x - 3)(4x + 12) = 0:
- Simplifying gives roots for \( x = \frac{1}{2} \) and \( x = -3 \), but not 12.
-
(x - 12)(x + 3) = 0 (same as before):
- This also matches perfectly, yielding the roots 12 and -3.
-
(x - 12)(x - 3) = 0:
- This gives the roots 12 and 3, not -3.
-
(4x + 2)(x + 3) = 0:
- This gives no roots of 12 and -3.
Based on the analysis:
The equations that have the solution set of {12, -3} are:
- (x - 12)(x + 3) = 0
- (x - 12)(x + 3) = 0 (duplicate; counted only once)
The selected equations are:
- (x - 12)(x + 3) = 0 (from both occurrences).
So the final answer is:
- \( (x - 12)(x + 3) = 0 \) (first occurrence)
- \( (x - 12)(x + 3) = 0 \) (second occurrence)