Problem 62 - Algebra 1

Question

The solution of a quadratic equation can be found by graphing each side separately and locating the points of intersection. You may wish to consult page 532 for help in approximating solutions. $$ -x^{2}-2=4 x^{2}+6 x-3 $$

Step-by-Step Solution

Answer

The solutions of the equation are \(x1 = \frac{-6 + \sqrt{56}}{10}\) and \(x2 = \frac{-6 - \sqrt{56}}{10}\).

Step 1: Bring All Terms to One Side of Equation

The equation can be brought into the standard form by bringing all terms to one side. So, -x^2 - 2 = 4x^2 + 6x - 3 becomes 0 = 5x^2 + 6x -1.

Step 2: Apply Quadratic Formula

The quadratic formula is defined as \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\) . For the quadratic equation \(5x^2+6x-1=0\), a = 5, b = 6, and c = -1. Substituting these values into the quadratic formula yields the solutions.

Step 3: Calculate Values of x

Using the quadratic formula, we get \(x = \frac{-6 \pm \sqrt{6^2 - 4*5*-1}}{2*5}\), which simplifies to \(x = \frac{-6 \pm \sqrt{36 + 20}}{10}\). Further simplification gives \(x = \frac{-6 \pm \sqrt{56}}{10}\). Therefore, we get \(x1 = \frac{-6 + \sqrt{56}}{10}\) and \(x2 = \frac{-6 - \sqrt{56}}{10}\).

Key Concepts

Quadratic Formula

The Quadratic Formula is a tool that helps us solve quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula is very helpful when factoring is either difficult or impossible.
It is defined as:

• \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)

To use the Quadratic Formula, we need to identify the coefficients \( a \), \( b \), and \( c \) of the quadratic equation. For example, if we have \( 5x^2 + 6x - 1 = 0 \), then \( a = 5 \), \( b = 6 \), and \( c = -1 \).
The discriminant, \( b^2 - 4ac \), is crucial because it determines the nature of the solutions:

• If \( b^2 - 4ac > 0 \), there are two real distinct solutions.
• If \( b^2 - 4ac = 0 \), there is one real solution.
• If \( b^2 - 4ac < 0 \), there are no real solutions, only complex ones.

So, when you apply the Quadratic Formula, it's important to calculate the discriminant first to know what to expect from your solutions.

Graphing Quadratic Functions

Graphing quadratic functions helps visualize the solutions of an equation. A quadratic function can be written in the form \( y = ax^2 + bx + c \), and its graph is a parabola.
Key elements of a quadratic graph include:

• **Vertex**: The highest or lowest point of the parabola.
• **Axis of symmetry**: A vertical line that divides the parabola into two mirror images, passing through the vertex.
• **Intercepts**: Points where the parabola intersects the x-axis (real solutions) and y-axis.

For the equation \( 5x^2 + 6x - 1 = 0 \), graphing involves setting \( f(x) = 5x^2 + 6x - 1 \) and finding the intercepts. The points where the graph meets the x-axis are the solutions of the equation. By doing this, students can visually confirm the solutions derived algebraically.
Understanding the shape and position of the parabola aids in comprehending the real-world context of where solutions occur, making it clearer how the solutions relate to practical problems.

Algebraic Solutions

Achieving algebraic solutions involves solving equations through manipulation of algebraic expressions. This process often starts with setting the equation to zero by collecting all terms on one side of the equation. This is essential to ensure the equation takes a standard quadratic form, \( ax^2 + bx + c = 0 \).
For example, converting the given equation \(-x^2 - 2 = 4x^2 + 6x - 3\) to \(0 = 5x^2 + 6x - 1\) highlights this step.

• Move all terms to one side to align with the quadratic equation format.

Once in standard form, methods such as factoring, completing the square, or using the Quadratic Formula are used.
The choice depends on the ease and the particular solutions anticipated. Quadratic Formula is often preferred for its universality and ability to always find solutions, as long as their computation is possible.