Which of the following quadratic equations have the solutions x = -3 and x = 5?
A. -1/2(x+3)(x-5)=0
B. (x+3)(x-5)=0
C. (x-3)(x+5)=0
D. -1/2(x-3)(x+5)=0
To determine which of the given quadratic equations have the solutions x = -3 and x = 5, we need to substitute these values into each equation and check if it equals zero.
Let's start with option A:
A. -1/2(x+3)(x-5)=0
Substituting x = -3 into the equation:
-1/2((-3)+3)((-3)-5) = 0
-1/2(0)(-8) = 0
0 = 0
The equation yields 0 when x = -3.
Now, let's substitute x = 5 into option A:
-1/2((5)+3)((5)-5) = 0
-1/2(8)(0) = 0
0 = 0
The equation also gives 0 when x = 5.
Therefore, option A has the solutions x = -3 and x = 5.
Now, let's continue with option B:
B. (x+3)(x-5)=0
Substituting x = -3:
((-3)+3)((-3)-5) = 0
0(-8) = 0
0 = 0
The equation yields 0 when x = -3.
Substituting x = 5:
((5)+3)((5)-5) = 0
8(0) = 0
0 = 0
The equation also gives 0 when x = 5.
Thus, option B has the solutions x = -3 and x = 5 as well.
Let's move on to option C:
C. (x-3)(x+5)=0
Substituting x = -3:
((-3)-3)((-3)+5) = 0
(-6)(2) = -12
-12 ≠ 0
The equation does not yield 0 when x = -3.
Substituting x = 5:
((5)-3)((5)+5) = 0
(2)(10) = 20
20 ≠ 0
The equation also does not give 0 when x = 5.
Therefore, option C does not have the solutions x = -3 and x = 5.
Finally, let's examine option D:
D. -1/2(x-3)(x+5)=0
Substituting x = -3:
-1/2((-3)-3)((-3)+5) = 0
-1/2(-6)(2) = -12
-12 ≠ 0
The equation does not yield 0 when x = -3.
Substituting x = 5:
-1/2((5)-3)((5)+5) = 0
-1/2(2)(10) = -20
-20 ≠ 0
The equation also does not give 0 when x = 5.
Therefore, option D does not have the solutions x = -3 and x = 5.
In conclusion, options A and B, which are -1/2(x+3)(x-5) and (x+3)(x-5), respectively, have the solutions x = -3 and x = 5.