Let's solve the given problem step-by-step.
The roots of the quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] are given as [tex]\( x = 2 \)[/tex] and [tex]\( x = -3 \)[/tex].
According to Vieta's formulas:
1. The sum of the roots ([tex]\( x_1 + x_2 \)[/tex]) is equal to [tex]\( -\frac{b}{a} \)[/tex].
2. The product of the roots ([tex]\( x_1 \cdot x_2 \)[/tex]) is equal to [tex]\( \frac{c}{a} \)[/tex].
Given:
[tex]\[ x_1 = 2 \][/tex]
[tex]\[ x_2 = -3 \][/tex]
First, let's calculate the sum of the roots:
[tex]\[ x_1 + x_2 = 2 + (-3) = -1 \][/tex]
So, we have:
[tex]\[ -\frac{b}{a} = -1 \][/tex]
[tex]\[ b = a \][/tex]
Next, let's calculate the product of the roots:
[tex]\[ x_1 \cdot x_2 = 2 \cdot (-3) = -6 \][/tex]
So, we have:
[tex]\[ \frac{c}{a} = -6 \][/tex]
[tex]\[ c = -6a \][/tex]
We assume [tex]\( a = 1 \)[/tex] for simplicity to find the relationships:
[tex]\[ b = a = 1 \][/tex]
[tex]\[ c = -6a = -6 \][/tex]
At this point, we have:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 1 \][/tex]
[tex]\[ c = -6 \][/tex]
Now, let's check which of the given options satisfies these values.
A) [tex]\( a + b + c = 1 + 1 - 6 = -4 \)[/tex] (This does not satisfy the equation.)
B) [tex]\( a - b + c = 1 - 1 - 6 = -6 \)[/tex] (This does not satisfy the equation.)
C) [tex]\( a - b - c = 1 - 1 + 6 = 6 \)[/tex] (This does not satisfy the equation.)
D) [tex]\( a + b - c = 1 + 1 + 6 = 8 \)[/tex] (This does not satisfy the equation.)
Since none of the above conditions [tex]\( a + b + c = 0 \)[/tex], [tex]\( a - b + c = 0 \)[/tex], [tex]\( a - b - c = 0 \)[/tex], [tex]\( a + b - c = 0 \)[/tex] exactly work with our computed values:
Therefore, the answer is:
[tex]\[ \text{None of the above} \][/tex]
