Answer :
To find the other possible solution for the quadratic equation [tex]\( x^2 + 8x + 15 = 0 \)[/tex], given that [tex]\( x = -5 \)[/tex] is one of the solutions, we can follow these steps:
1. Identify the standard form of the quadratic equation:
The given equation is [tex]\( x^2 + 8x + 15 = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 15 \)[/tex].
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 15 \)[/tex]:
[tex]\[ \Delta = 8^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4 \][/tex]
3. Use the quadratic formula to find the solutions:
The quadratic formula for the solutions [tex]\( x \)[/tex] of [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\(\Delta = 4\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{4}}{2 \cdot 1} = \frac{-8 \pm 2}{2} \][/tex]
4. Find the two possible values of [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-8 + 2}{2} = \frac{-6}{2} = -3 \][/tex]
[tex]\[ x_2 = \frac{-8 - 2}{2} = \frac{-10}{2} = -5 \][/tex]
Given that one of the solutions is [tex]\( x = -5 \)[/tex], the other solution must be:
[tex]\[ x = -3 \][/tex]
Thus, the other value of [tex]\( x \)[/tex] for the quadratic equation [tex]\( x^2 + 8x + 15 = 0 \)[/tex] is [tex]\(-3\)[/tex].
The correct answer is (C) -3.
1. Identify the standard form of the quadratic equation:
The given equation is [tex]\( x^2 + 8x + 15 = 0 \)[/tex].
Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 15 \)[/tex].
2. Calculate the discriminant:
The discriminant [tex]\(\Delta\)[/tex] of a quadratic equation [tex]\( ax^2 + bx + c = 0 \)[/tex] is given by:
[tex]\[ \Delta = b^2 - 4ac \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\( c = 15 \)[/tex]:
[tex]\[ \Delta = 8^2 - 4 \cdot 1 \cdot 15 = 64 - 60 = 4 \][/tex]
3. Use the quadratic formula to find the solutions:
The quadratic formula for the solutions [tex]\( x \)[/tex] of [tex]\( ax^2 + bx + c = 0 \)[/tex] is:
[tex]\[ x = \frac{-b \pm \sqrt{\Delta}}{2a} \][/tex]
Substituting [tex]\( a = 1 \)[/tex], [tex]\( b = 8 \)[/tex], and [tex]\(\Delta = 4\)[/tex]:
[tex]\[ x = \frac{-8 \pm \sqrt{4}}{2 \cdot 1} = \frac{-8 \pm 2}{2} \][/tex]
4. Find the two possible values of [tex]\( x \)[/tex]:
[tex]\[ x_1 = \frac{-8 + 2}{2} = \frac{-6}{2} = -3 \][/tex]
[tex]\[ x_2 = \frac{-8 - 2}{2} = \frac{-10}{2} = -5 \][/tex]
Given that one of the solutions is [tex]\( x = -5 \)[/tex], the other solution must be:
[tex]\[ x = -3 \][/tex]
Thus, the other value of [tex]\( x \)[/tex] for the quadratic equation [tex]\( x^2 + 8x + 15 = 0 \)[/tex] is [tex]\(-3\)[/tex].
The correct answer is (C) -3.
