Problem 59 - 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. $$ 3 x^{2}+2 x+5=6 x^{2} $$

Step-by-Step Solution

Answer

The solutions to the equation are \(x=-1\) and \(x=5/3\).

Step 1: Simplify the equation

First, you need to simplify the equation by setting it equal to zero. This is done by subtracting all terms from right side of the equation from left side: \(3x^2+2x+5-6x^2=0\), which simplifies to \(-3x^2+2x+5=0\).

Step 2: Identify values of a, b and c

Now, you need to identify the values of a, b and c. From the simplified equation \(-3x^2+2x+5=0\), a=-3, b=2, c=5.

Step 3: Substitute values into the quadratic formula

Insert the values of a, b and c into the quadratic formula. This gives: \(x=\frac{-b±\sqrt{b^2 -4ac}}{2a} = \frac{-2±\sqrt{(2)^2 -4*(-3)*5}}{2*(-3)}\).

Step 4: Simplify the solution

Upon simplifying the equation, you find: \(x=\frac{-2±\sqrt{4+60}}{-6} = \frac{-2±\sqrt{64}}{-6} = \frac{-2±8}{-6}\). This further simplifies to two solutions: \(x=\frac{-2+8}{-6}=-1\) and \(x=\frac{-2-8}{-6}=5/3\).

Key Concepts

Quadratic Formula

Quadratic equations, represented in the standard form as \(ax^2 + bx + c = 0\), are equations where the highest degree of the variable \(x\) is two. The quadratic formula is a powerful tool for solving these equations and is given by: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula can be used to find the roots (or solutions) of the equation, \(x\), where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. Let's break down the formula:

• The term \(b^2 - 4ac\) is known as the "discriminant." It determines the nature of the roots.
• If the discriminant is positive, the equation has two real roots.
• If the discriminant is zero, there is one real root.
• If the discriminant is negative, the equation has two complex roots.

In our problem, by identifying \(a = -3\), \(b = 2\), and \(c = 5\), we substitute these into the quadratic formula to solve for \(x\). This systematic approach ensures that we can attentively evaluate different scenarios of \(x\), providing us with the complete solutions (in this case, \(x = -1\) and \(x = \frac{5}{3}\)).

Graphing Methods

Graphing is a visual method used to find the solutions (or roots) of a quadratic equation. By plotting the equation on a coordinate plane, we can visually identify where it intersects the x-axis. These points of intersection are the solutions of the equation.
To graph a quadratic equation:

• Rewrite the equation in the form \(y = ax^2 + bx + c\) if necessary. For our example, this starts with simplifying the two sides before graphing one: \(3x^2 + 2x + 5 = 6x^2\) which simplifies to \(-3x^2 + 2x + 5 = 0\).
• Identify the vertex and select points on either side to form a parabola.
• Determine the axis of symmetry, which is the line \(x = -\frac{b}{2a}\).
• Calculate and plot key points and sketch the curve.
• Find where the parabola intersects the x-axis. These are the solutions \(x\).

Graphing provides a tangible way to view the solutions. In our problem, the values \(x = -1 \) and \(x = \frac{5}{3}\) would appear where the graph intersects the x-axis.

Algebraic Solutions

Finding solutions to quadratic equations involves diverse methods, and among the most crucial algebraic techniques are simplifying the equation and factoring. In algebra, simplifying an equation generally means performing operations such as combining like terms and moving all terms to one side to form zero on the other side.
The step-by-step process generally involves:

• Rearranging terms on one side of the equation for simplification.
• Identifying coefficients \(a\), \(b\), \(c\) from the standard quadratic form.
• Attempting to factor the quadratic expression or applying the quadratic formula, if factoring seems complicated.

In the given exercise, we first simplify \(3x^2 + 2x + 5 - 6x^2 = 0\) to \(-3x^2 + 2x + 5 = 0\). From here, identifying \(a\), \(b\), and \(c\) allows for direct application of the quadratic formula, providing precise solutions when graphical methods or simple factoring are not enough. Thanks to these algebraic solutions, one can achieve exact roots regardless of the equation’s complexity.