Problem 24
Question
Solve each equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. $$ 14 x+x^{2}+49=0 $$
Step-by-Step Solution
Verified Answer
The equation \( 14x + x^2 + 49 = 0 \) has one repeated root at \( x = -7 \).
1Step 1: Rearrange the Equation
The given equation is already arranged, but let's write it in standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 14 \), and \( c = 49 \). Hence, the equation is \( x^2 + 14x + 49 = 0 \).
2Step 2: Identify the Function to Graph
The function to graph is the quadratic function \( f(x) = x^2 + 14x + 49 \). This will help us find the values of \( x \) where \( f(x) = 0 \).
3Step 3: Determine the Vertex of the Parabola
The vertex of a parabola in the form \( y = ax^2 + bx + c \) can be found using \( x = -\frac{b}{2a} \). Here, \( a = 1 \) and \( b = 14 \), so \( x = -\frac{14}{2} = -7 \). Calculate \( y \) using \( y = (-7)^2 + 14(-7) + 49 = 0 \). So, the vertex \((-7, 0)\) is on the x-axis.
4Step 4: Plot the Parabola
Using a graphing tool, plot the quadratic function \( f(x) = x^2 + 14x + 49 \). Since the vertex of the parabola is \((-7, 0)\) and \( f(x) \) is a parabola opening upwards, note that it touches the x-axis at \( x = -7 \), indicating a repeated root.
5Step 5: Conclusion from the Graph
From the graph, observe that the parabola only touches the x-axis at \( x = -7 \). Therefore, there is only one root, and it is exact and repeated.
Key Concepts
Graphing MethodStandard FormVertex of a ParabolaRepeated Root
Graphing Method
Graphing is an effective way to visually solve quadratic equations. To utilize the graphing method, you first transform the quadratic equation into a function format. By plotting this function on a graph, you can observe where it intersects the x-axis, if at all. These intersection points are the solutions or "roots" of the equation. For example, with the equation given, we graph the function \( f(x) = x^2 + 14x + 49 \). By observing the graph, we see that it touches the x-axis at the point corresponding to the root of the equation.
- To graph efficiently, identify key points such as the vertex and y-intercept.
- The graph should be a smooth curve, opening upwards if the coefficient of \( x^2 \) is positive, or downwards if negative.
Standard Form
In algebra, the standard form of a quadratic equation is crucial for solving and interpreting. Standard form is expressed as \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants.
- Consistently automating the equation into this form simplifies many solving methods, like factoring or using the quadratic formula.
- It also aids in identifying key features of the parabola, such as the direction it opens.
Vertex of a Parabola
The vertex of a parabola is a vital point that represents the maximum or minimum value of the function. It serves as the turning point of the graph. To find the vertex's x-coordinate in a quadratic equation in standard form, use the formula\[ x = -\frac{b}{2a} \]. Using our example, with \( a = 1 \) and \( b = 14 \), the x-coordinate is calculated as \(-7\). To find the y-coordinate, substitute \( x = -7 \) back into the equation: \( y = (-7)^2 + 14(-7) + 49 = 0 \). Thus, the vertex \((-7, 0)\) indicates that the parabola merely touches the x-axis, pinpointing a repeated root. This point is key to graphing and understanding the symmetry of the quadratic function.
- The vertex also helps to determine if the parabola has a maximum or minimum point, crucial for optimization problems.
Repeated Root
A repeated root in a quadratic equation occurs when the parabola touches the x-axis at exactly one point. This means the root is a "double root", showing in the solution of the equation as a single repeated value. In our example: \( x^2 + 14x + 49 = 0 \), the graph intersects the x-axis at \( x = -7 \). Since the parabola only touches the axis at this point, \( x = -7 \) becomes the repeated root.
- This situation occurs when the discriminant \(( b^2 - 4ac)\) equals zero. For our equation, \( (14)^2 - 4(1)(49) = 0 \).
- The graph verifies this as there is only one solution due to the tangent point.
Other exercises in this chapter
Problem 24
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Solve each equation by using the method of your choice. Find exact solutions. \(x^{2}-30 x-64=0\)
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