To solve the equation \(x^2 + 2x = 3\), we can first rewrite it in standard quadratic form:
\[ x^2 + 2x - 3 = 0 \]
Next, we'll create a function from the left side of the equation:
\[ f(x) = x^2 + 2x - 3 \]
We'll find the solutions to the equation by determining where the graph of the function \(f(x)\) intersects the x-axis (where \(f(x) = 0\)).
To sketch the graph of \(f(x)\), we can find the vertex and the y-intercept.
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Finding the Vertex: The vertex of a parabola \(ax^2 + bx + c\) is given by the x-coordinate \(x = -\frac{b}{2a}\):
- Here, \(a = 1\), \(b = 2\) so: \[ x = -\frac{2}{2 \cdot 1} = -1 \] To find the y-coordinate of the vertex, substitute \(x = -1\) into the function: \[ f(-1) = (-1)^2 + 2(-1) - 3 = 1 - 2 - 3 = -4 \] The vertex is at the point \((-1, -4)\).
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Finding the Y-Intercept: The y-intercept occurs when \(x = 0\): \[ f(0) = 0^2 + 2(0) - 3 = -3 \] The y-intercept point is \((0, -3)\).
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Finding the X-Intercepts: We can find the x-intercepts by factoring the quadratic equation \(x^2 + 2x - 3\): \[ (x + 3)(x - 1) = 0 \] Setting each factor to zero gives us: \[ x + 3 = 0 \Rightarrow x = -3 \] \[ x - 1 = 0 \Rightarrow x = 1 \]
The x-intercepts are at the points \((-3, 0)\) and \((1, 0)\).
Summary of the Solutions
The solutions to the original equation \(x^2 + 2x = 3\) are where \(f(x) = 0\):
- \(x = -3\)
- \(x = 1\)
Thus, the correct response is: x = −3, 1
If providing this in a multiple-choice format:
- x = −3, 1 (correct response)
- x = 0, 1
- x = 2, 0
- no solution
Conclusion
The solutions to the equation \(x^2 + 2x = 3\) are \(x = -3\) and \(x = 1\).