Problem 26 - Algebra 2

Question

Solve each equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. $$ x^{2}+2 x+5=0 $$

Step-by-Step Solution

Answer

The equation has no real roots; therefore, no x-intersections exist.

Step 1: Understand the Problem

To find the roots of the equation \( x^2 + 2x + 5 = 0 \) by graphing, we need to graph the function \( y = x^2 + 2x + 5 \) and look for the x-values where the graph intersects the x-axis. These are the roots of the equation.

Step 2: Determine the Vertex of the Parabola

The function \( y = x^2 + 2x + 5 \) is a quadratic function. The vertex form of a quadratic function is \( y = a(x-h)^2 + k \). We can find the vertex of this function by using the formula \(-\frac{b}{2a}\) to find the x-coordinate of the vertex. For our equation, \( a = 1 \) and \( b = 2 \), so the x-coordinate of the vertex is \(-\frac{2}{2(1)} = -1\). The y-coordinate is found by substituting \( -1 \) into the function: \( y = (-1)^2 + 2(-1) + 5 = 4 \). Thus, the vertex is \((-1, 4)\).

Step 3: Plot the Vertex and Additional Points

Plot the vertex \((-1, 4)\) on the graph. Then, calculate a few additional points by choosing x-values around the vertex. For example: \( x = -2, y = (-2)^2 + 2(-2) + 5 = 5 \); \( x = 0, y = 0^2 + 2(0) + 5 = 5 \). Plot these points: \((-2, 5)\) and \((0, 5)\).

Step 4: Sketch the Parabola

Using the points \((-1, 4)\), \((-2, 5)\), and \((0, 5)\), sketch a parabola. The shape will open upwards because the coefficient of \( x^2 \) is positive.

Step 5: Analyze the Graph

Observe that the parabola does not intersect the x-axis. Since there are no x-intersections, it indicates that there are no real roots for the equation \( x^2 + 2x + 5 = 0 \). This means the roots are complex numbers.

Key Concepts

Graphing Quadratics

Graphing quadratic equations involves plotting their functions on a coordinate plane. A quadratic equation takes the form \( ax^2 + bx + c = 0 \). When graphed, it forms a U-shaped curve known as a parabola. The way the parabola opens depends on the sign of the coefficient \( a \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), the parabola opens downwards.

In this case, for the equation \( y = x^2 + 2x + 5 \), we plot the equation on the graph. Choose a series of \( x \) values and compute the corresponding \( y \) values, as shown in the example provided. Plot these points and sketch a smooth curve through them to visualize the parabola.
This visualization helps us to determine the nature of the roots. If the parabola touches or crosses the x-axis, the roots are real. If it does not, the roots are complex.

Vertex of Parabola

The vertex of a parabola is its highest or lowest point, depending on whether it opens down or up. For a quadratic equation in standard form \( y = ax^2 + bx + c \), you can determine the vertex using the formula for the x-coordinate as \(-\frac{b}{2a}\). This gives the x position within the vertex.
For the equation \( y = x^2 + 2x + 5 \), we find the vertex by calculating:

• The x-coordinate: \(-\frac{2}{2 \times 1} = -1\)
• The y-coordinate by substituting \( x = -1 \) back into the function: \( y = (-1)^2 + 2(-1) + 5 = 4 \)

The vertex is \((-1, 4)\). Knowing this point is crucial, as it helps when sketching the parabola and understanding its general shape and position on the graph.

Complex Roots of Polynomials

A quadratic equation may not always have real roots. When the parabola does not intersect with the x-axis, the equation has complex roots. Complex roots occur when the discriminant of the quadratic equation, known by the formula \( b^2 - 4ac \), is negative.
For the equation \( x^2 + 2x + 5 = 0 \), calculate the discriminant:

• Given \( a = 1 \), \( b = 2 \), and \( c = 5 \)
• Discriminant: \( 2^2 - 4 \times 1 \times 5 = 4 - 20 = -16 \)

A negative discriminant indicates non-real solutions, specifically complex roots. Complex numbers are of the form \( a + bi \), where \( i \) is the imaginary unit \( \sqrt{-1} \). Therefore, when graphing doesn’t show any x-intersections, it’s important to consider complex roots as the solution to the equation.