Problem 39 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ 3 x^{2}+6 x-5=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = \frac{-3 + 2\sqrt{6}}{3} \) and \( x = \frac{-3 - 2\sqrt{6}}{3} \).

Step 1: Identify Coefficients

First, we need to identify the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \). In this equation, we have \( a = 3 \), \( b = 6 \), and \( c = -5 \).

Step 2: Apply the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We'll use this formula to find the solutions for \( x \).

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \) to determine the nature of the roots. Substitute \( b = 6 \), \( a = 3 \), and \( c = -5 \): \( 6^2 - 4 \cdot 3 \cdot (-5) = 36 + 60 = 96 \). The discriminant is 96, which is positive, indicating two distinct real solutions.

Step 4: Compute the Square Root of the Discriminant

Calculate \( \sqrt{96} \). Simplifying, we have \( \sqrt{96} = \sqrt{16 \cdot 6} = 4\sqrt{6} \).

Step 5: Find the Two Solutions

Plug the values back into the quadratic formula: \( x = \frac{-6 \pm 4\sqrt{6}}{6} \). Simplifying further, we find the two solutions: \( x_1 = \frac{-6 + 4\sqrt{6}}{6} = \frac{-3 + 2\sqrt{6}}{3} \) and \( x_2 = \frac{-6 - 4\sqrt{6}}{6} = \frac{-3 - 2\sqrt{6}}{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that helps us find the solutions of a quadratic equation. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are the coefficients. The quadratic formula is:

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

This formula enables us to calculate the zeros of the quadratic equation.
The important part is to correctly substitute the coefficients \( a \), \( b \), and \( c \) into the formula. The \( \pm \) sign indicates there will be two potential solutions once evaluated.

Discriminant of Quadratic Equation

The discriminant is a part of the quadratic formula that helps us understand the nature of the roots of the quadratic equation. It is found in the formula under the square root: \( b^2 - 4ac \).
The value of the discriminant gives us crucial information:

• If the discriminant is positive, there are two real and distinct solutions.
• If it is zero, there is exactly one real solution, also known as a repeated or double root.
• If the discriminant is negative, there are no real solutions, but two complex solutions.

Knowing the value of the discriminant helps us predict the type of solutions without actually solving the entire equation.

Real Solutions of Quadratic Equations

Real solutions of quadratic equations are the values of \( x \) that satisfy the equation \( ax^2 + bx + c = 0 \) and are real numbers rather than complex numbers.
For our specific equation, with the coefficients \( a = 3 \), \( b = 6 \), and \( c = -5 \), we calculated the discriminant to be 96. This positive value indicates that there are two distinct real solutions.
Upon applying the quadratic formula, we find these solutions:

• \( x_1 = \frac{-3 + 2\sqrt{6}}{3} \)
• \( x_2 = \frac{-3 - 2\sqrt{6}}{3} \)

These are the points where the quadratic equation crosses the x-axis on a graph. Understanding these solutions gives insight into the behavior of the quadratic function across different values of \( x \).