complete the square to identify the two values that solve the following quadratic equation: x^2 + 8x = 9.

Answer 1

To solve the quadratic equation \(x^2 + 8x = 9\) by completing the square, follow these steps:

1. Move the constant to the other side: \[ x^2 + 8x - 9 = 0 \] Now, we isolate the quadratic and linear terms: \[ x^2 + 8x = 9 \]

2. Complete the square: Take the coefficient of \(x\) (which is 8), divide it by 2, and square it: \[ \left(\frac{8}{2}\right)^2 = 4^2 = 16 \]

3. Add and subtract this square to the left side of the equation: \[ x^2 + 8x + 16 - 16 = 9 \] This can be rewritten as: \[ (x + 4)^2 - 16 = 9 \]

4. Move the subtracted square back to the right: \[ (x + 4)^2 = 9 + 16 \] \[ (x + 4)^2 = 25 \]

5. Take the square root of both sides: \[ x + 4 = \pm 5 \]

6. Solve for \(x\):

   • For the positive square root: \[ x + 4 = 5 \quad \Rightarrow \quad x = 5 - 4 \quad \Rightarrow \quad x = 1 \]
   • For the negative square root: \[ x + 4 = -5 \quad \Rightarrow \quad x = -5 - 4 \quad \Rightarrow \quad x = -9 \]

Thus, the two values that solve the equation \(x^2 + 8x = 9\) are: \[ x = 1 \quad \text{and} \quad x = -9 \]

Answer 2

no it isn't it is just one and it is nor -9 or 1

Answer 3

I apologize for the confusion. Let's carefully go through the steps to complete the square for the given equation \(x^2 + 8x = 9\) again to find the correct solutions.

1. Start with the original equation: \[ x^2 + 8x = 9 \]

2. Move the constant to the other side: \[ x^2 + 8x - 9 = 0 \]

3. Complete the square on the left side:

   • Take the coefficient of \(x\) (which is 8), divide it by 2, and square it: \[ \left(\frac{8}{2}\right)^2 = 4^2 = 16 \]
   • Add and subtract this square inside the equation: \[ x^2 + 8x + 16 - 16 = 9 \] This gives us: \[ (x + 4)^2 - 16 = 9 \] Now add 16 to both sides: \[ (x + 4)^2 = 25 \]

4. Take the square root of both sides: \[ x + 4 = \pm 5 \]

5. Solve for \(x\):

   • For the positive root: \[ x + 4 = 5 \quad \Rightarrow \quad x = 1 \]
   • For the negative root: \[ x + 4 = -5 \quad \Rightarrow \quad x = -9 \]

Therefore, the solutions are: \[ x = 1 \quad \text{and} \quad x = -9 \]

If you meant that there is only one solution, we can check the discriminant of the equation to confirm whether it's a perfect square or if there's any other nuance to the problem.

The discriminant of the quadratic equation \(ax^2 + bx + c = 0\) is given by \(D = b^2 - 4ac\).

For our equation \(x^2 + 8x - 9 = 0\):

• \(a = 1\)
• \(b = 8\)
• \(c = -9\)

Calculating the discriminant: \[ D = 8^2 - 4 \cdot 1 \cdot (-9) = 64 + 36 = 100 \]

Since the discriminant is positive, there are indeed two distinct real roots.

If there's something specific that might have led to the misunderstanding, please let me know!