Answer :
To determine whether [tex]\( x = -9 \)[/tex] and [tex]\( x = -3 \)[/tex] are indeed the solutions to the quadratic equation [tex]\( x^2 + 12x + 27 = 0 \)[/tex], let's carefully go through the steps involved in solving a quadratic equation.
### Step 1: Identify the quadratic equation
The given quadratic equation is:
[tex]\[ x^2 + 12x + 27 = 0 \][/tex]
### Step 2: Use the quadratic formula
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
From the given equation [tex]\( x^2 + 12x + 27 = 0 \)[/tex], we can identify:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 12 \][/tex]
[tex]\[ c = 27 \][/tex]
### Step 3: Calculate the discriminant
The discriminant ([tex]\(\Delta\)[/tex]) is given by [tex]\( b^2 - 4ac \)[/tex].
[tex]\[ \Delta = 12^2 - 4(1)(27) = 144 - 108 = 36 \][/tex]
### Step 4: Compute the solutions using the quadratic formula
Now substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant back into the quadratic formula:
[tex]\[ x = \frac{-12 \pm \sqrt{36}}{2(1)} \][/tex]
[tex]\[ x = \frac{-12 \pm 6}{2} \][/tex]
This gives us two possible solutions:
[tex]\[ x_1 = \frac{-12 + 6}{2} = \frac{-6}{2} = -3 \][/tex]
[tex]\[ x_2 = \frac{-12 - 6}{2} = \frac{-18}{2} = -9 \][/tex]
### Step 5: Verify the solutions
We found two solutions:
[tex]\[ x = -3 \][/tex]
[tex]\[ x = -9 \][/tex]
These values can be verified by substituting them back into the original quadratic equation to ensure that both satisfy [tex]\( x^2 + 12x + 27 = 0 \)[/tex].
For [tex]\( x = -3 \)[/tex]:
[tex]\[ (-3)^2 + 12(-3) + 27 = 9 - 36 + 27 = 0 \][/tex]
For [tex]\( x = -9 \)[/tex]:
[tex]\[ (-9)^2 + 12(-9) + 27 = 81 - 108 + 27 = 0 \][/tex]
### Conclusion
Both [tex]\( x = -3 \)[/tex] and [tex]\( x = -9 \)[/tex] satisfy the original quadratic equation. Therefore, I agree with the solutions presented by your classmate. Both [tex]\( x = -9 \)[/tex] and [tex]\( x = -3 \)[/tex] are the correct [tex]\( x \)[/tex]-intercepts of the quadratic function [tex]\( x^2 + 12x + 27 \)[/tex].
### Step 1: Identify the quadratic equation
The given quadratic equation is:
[tex]\[ x^2 + 12x + 27 = 0 \][/tex]
### Step 2: Use the quadratic formula
The quadratic formula is:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex] are the coefficients of the equation [tex]\( ax^2 + bx + c = 0 \)[/tex].
From the given equation [tex]\( x^2 + 12x + 27 = 0 \)[/tex], we can identify:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 12 \][/tex]
[tex]\[ c = 27 \][/tex]
### Step 3: Calculate the discriminant
The discriminant ([tex]\(\Delta\)[/tex]) is given by [tex]\( b^2 - 4ac \)[/tex].
[tex]\[ \Delta = 12^2 - 4(1)(27) = 144 - 108 = 36 \][/tex]
### Step 4: Compute the solutions using the quadratic formula
Now substitute [tex]\( a \)[/tex], [tex]\( b \)[/tex], and the discriminant back into the quadratic formula:
[tex]\[ x = \frac{-12 \pm \sqrt{36}}{2(1)} \][/tex]
[tex]\[ x = \frac{-12 \pm 6}{2} \][/tex]
This gives us two possible solutions:
[tex]\[ x_1 = \frac{-12 + 6}{2} = \frac{-6}{2} = -3 \][/tex]
[tex]\[ x_2 = \frac{-12 - 6}{2} = \frac{-18}{2} = -9 \][/tex]
### Step 5: Verify the solutions
We found two solutions:
[tex]\[ x = -3 \][/tex]
[tex]\[ x = -9 \][/tex]
These values can be verified by substituting them back into the original quadratic equation to ensure that both satisfy [tex]\( x^2 + 12x + 27 = 0 \)[/tex].
For [tex]\( x = -3 \)[/tex]:
[tex]\[ (-3)^2 + 12(-3) + 27 = 9 - 36 + 27 = 0 \][/tex]
For [tex]\( x = -9 \)[/tex]:
[tex]\[ (-9)^2 + 12(-9) + 27 = 81 - 108 + 27 = 0 \][/tex]
### Conclusion
Both [tex]\( x = -3 \)[/tex] and [tex]\( x = -9 \)[/tex] satisfy the original quadratic equation. Therefore, I agree with the solutions presented by your classmate. Both [tex]\( x = -9 \)[/tex] and [tex]\( x = -3 \)[/tex] are the correct [tex]\( x \)[/tex]-intercepts of the quadratic function [tex]\( x^2 + 12x + 27 \)[/tex].
