Question
Solve the equation by graphing the related function.
x squared plus 2x2+2equals=0
x squared plus 2x2+2equals=0
Answers
GPT-4o mini
To solve the equation \( x^2 + 2x + 2 = 0 \) by graphing, we'll first consider the function associated with the equation:
\[
f(x) = x^2 + 2x + 2
\]
### Step 1: Identify the Shape of the Graph
The function \( f(x) \) is a quadratic function, and its graph will be a parabola. Since the coefficient of \( x^2 \) is positive, the parabola opens upward.
### Step 2: Find the Vertex
The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 2 \).
\[
x = -\frac{2}{2 \cdot 1} = -1
\]
Now, substitute \( x = -1 \) to find the corresponding \( y \)-value:
\[
f(-1) = (-1)^2 + 2(-1) + 2 = 1 - 2 + 2 = 1
\]
So the vertex is at the point \((-1, 1)\).
### Step 3: Determine the Discriminant
To find the roots of the quadratic equation, we can compute the discriminant \( D = b^2 - 4ac \):
\[
D = 2^2 - 4(1)(2) = 4 - 8 = -4
\]
Since the discriminant is negative, there are no real roots. Therefore, the graph of the function will not intersect the x-axis.
### Step 4: Graph the Function
You can visualize the function \( f(x) = x^2 + 2x + 2 \) as a parabola opening upwards with vertex at \((-1, 1)\).
### Conclusion
The equation \( x^2 + 2x + 2 = 0 \) has no real solutions, as the graph does not intersect the x-axis. The only solutions are complex. To find them, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-4}}{2 \cdot 1} = \frac{-2 \pm 2i}{2} = -1 \pm i
\]
Thus, the solutions are:
\[
x = -1 + i \quad \text{and} \quad x = -1 - i
\]
\[
f(x) = x^2 + 2x + 2
\]
### Step 1: Identify the Shape of the Graph
The function \( f(x) \) is a quadratic function, and its graph will be a parabola. Since the coefficient of \( x^2 \) is positive, the parabola opens upward.
### Step 2: Find the Vertex
The vertex of the parabola can be found using the formula \( x = -\frac{b}{2a} \), where \( a = 1 \) and \( b = 2 \).
\[
x = -\frac{2}{2 \cdot 1} = -1
\]
Now, substitute \( x = -1 \) to find the corresponding \( y \)-value:
\[
f(-1) = (-1)^2 + 2(-1) + 2 = 1 - 2 + 2 = 1
\]
So the vertex is at the point \((-1, 1)\).
### Step 3: Determine the Discriminant
To find the roots of the quadratic equation, we can compute the discriminant \( D = b^2 - 4ac \):
\[
D = 2^2 - 4(1)(2) = 4 - 8 = -4
\]
Since the discriminant is negative, there are no real roots. Therefore, the graph of the function will not intersect the x-axis.
### Step 4: Graph the Function
You can visualize the function \( f(x) = x^2 + 2x + 2 \) as a parabola opening upwards with vertex at \((-1, 1)\).
### Conclusion
The equation \( x^2 + 2x + 2 = 0 \) has no real solutions, as the graph does not intersect the x-axis. The only solutions are complex. To find them, we can use the quadratic formula:
\[
x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-2 \pm \sqrt{-4}}{2 \cdot 1} = \frac{-2 \pm 2i}{2} = -1 \pm i
\]
Thus, the solutions are:
\[
x = -1 + i \quad \text{and} \quad x = -1 - i
\]