Problem 42 - Algebra for College Students

Question

Solve each quadratic equation using the method that seems most appropriate. $$3 x^{2}+6 x=1$$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{-3 + 2\sqrt{3}}{3} \) and \( x = \frac{-3 - 2\sqrt{3}}{3} \).

Step 1: Write the equation in standard form

First, let's rewrite the equation so it is in the standard form of a quadratic equation, which is \( ax^2 + bx + c = 0 \). To do this, subtract 1 from both sides of the equation: \[ 3x^2 + 6x - 1 = 0 \] Now, the quadratic equation is in standard form with \( a = 3 \), \( b = 6 \), and \( c = -1 \).

Step 2: Identify suitable method for solving

Since the equation is a quadratic in standard form, you can use various methods such as factoring, completing the square, or the quadratic formula. Factoring isn't straightforward due to non-integer solutions, so it's best to use the quadratic formula.

Step 3: Apply the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute in the values \( a = 3 \), \( b = 6 \), and \( c = -1 \).

Step 4: Calculate the discriminant

Calculate the discriminant, \( b^2 - 4ac \): \[ b^2 - 4ac = 6^2 - 4 \cdot 3 \cdot (-1) = 36 + 12 = 48 \] Since the discriminant is positive, there are two real solutions.

Step 5: Solve for x using the quadratic formula

Substitute the discriminant and the constants into the quadratic formula: \[ x = \frac{-6 \pm \sqrt{48}}{6} \] The square root of 48 simplifies: \[ \sqrt{48} = 4\sqrt{3} \] Thus, the expression becomes: \[ x = \frac{-6 \pm 4\sqrt{3}}{6} \] This simplifies further to: \[ x = \frac{-3 \pm 2\sqrt{3}}{3} \]

Step 6: Write the solutions

The solutions to the equation \( 3x^2 + 6x - 1 = 0 \) are: \[ x = \frac{-3 + 2\sqrt{3}}{3} \] \[ x = \frac{-3 - 2\sqrt{3}}{3} \]

Key Concepts

Quadratic Formula

The quadratic formula is your go-to tool when you want to solve any quadratic equation of the form \( ax^2 + bx + c = 0 \). This formula is especially handy when factoring is tough or impossible. The formula is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here's how to use it:

• Identify your coefficients: \( a \), \( b \), and \( c \) from the equation's standard form.
• Calculate the discriminant \((b^2 - 4ac)\).
• Substitute \( a \), \( b \), and the discriminant into the formula.
• Solve for \( x \) to find your solutions.

With this, solving a quadratic equation becomes straightforward!

Standard Form

Before you can apply the quadratic formula, the equation needs to be in the standard form: \( ax^2 + bx + c = 0 \). This format is essential because it helps in identifying the coefficients necessary for the formula.Look at \( 3x^2 + 6x - 1 = 0 \), the equation is now in standard form where:

• \( a = 3 \)
• \( b = 6 \)
• \( c = -1 \)

Ensure all terms are on one side and the equation equals zero. Then, you're set to use formulas or other methods to solve it.

Discriminant

The discriminant is a powerful component of the quadratic formula. It tells you the nature of the solutions you can expect. The discriminant is calculated as \( b^2 - 4ac \). For the equation \( 3x^2 + 6x - 1 = 0 \), the discriminant is:
\[ 6^2 - 4 \cdot 3 \cdot (-1) = 36 + 12 = 48 \]
The sign of the discriminant tells a lot:

• If it's positive, like in our case (48), expect two distinct real solutions.
• If it's zero, expect one real solution, a perfect square situation.
• If negative, the solutions will be complex or imaginary.

This insight is crucial for anticipating the kind of results your quadratic formula will yield.

Real Solutions

In the world of quadratic equations, real solutions are the real numbers that satisfy the equation. When using the quadratic formula, you can identify the solutions as real based on the discriminant.Since our discriminant was 48 (a positive number), the equation \( 3x^2 + 6x - 1 = 0 \) has two real solutions. These solutions can be expressed as:
\[ x = \frac{-3 + 2\sqrt{3}}{3} \] and\[ x = \frac{-3 - 2\sqrt{3}}{3} \]Real solutions indicate where the graph of the quadratic equation intersects the x-axis. Each distinct real solution corresponds to a different point of intersection, or root, making them significant in understanding the equation's behavior.