Help me please.
Solve the equation:
(x-3)(x+2)=0
Enter the solution with the lowest value.
you know that if the product is zero, one of the factors must be zero. SO, you have
x-3 = 0
or
x+2 = 0
The solutions are therefore x = -2 or x = 3.
I assume you can now pick the solution with the lower value. Don't forget your algebra I, now that you're taking pre-calc!
oh i see, how could i have forgotten!!
To solve the equation (x-3)(x+2)=0, we can set each factor equal to zero and solve for x.
Setting x-3=0 gives x=3.
Setting x+2=0 gives x=-2.
Therefore, the solutions to the equation are x=3 and x=-2.
Since we need to enter the solution with the lowest value, the lowest value solution is x=-2.
To solve the equation (x-3)(x+2) = 0, you need to find the values of x that make the left side of the equation equal to zero.
To do this, you can set each factor equal to zero and solve for x separately.
First, set (x-3) = 0:
x - 3 = 0
Adding 3 to both sides of the equation:
x = 3
Next, set (x+2) = 0:
x + 2 = 0
Subtracting 2 from both sides of the equation:
x = -2
So, the values of x that satisfy the equation are x = 3 and x = -2.
Now, to find the solution with the lowest value, you compare the two values.
Since -2 is smaller than 3, the solution with the lowest value is x = -2.