91Ó°ÊÓ

The quadratic formula is a powerful tool used to find the solutions of a quadratic equation, which typically has the form \( ax^2 + bx + c = 0 \). To use the quadratic formula, we need to identify the coefficients: \( a \), \( b \), and \( c \). In the given equation \( 3x^2 + 6x - 5 = 0 \), these coefficients are \( a = 3 \), \( b = 6 \), and \( c = -5 \).
The formula itself is expressed as:

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

Plugging in our coefficients into this formula allows us to find the roots of the quadratic equation. One key aspect of this formula is the section under the square root sign, known as the discriminant, which provides insight into the nature of the solutions.

The discriminant is a component of the quadratic formula located under the square root sign, represented as \( b^2 - 4ac \). It plays a crucial role in determining the nature and number of solutions for a quadratic equation.
In our example, the discriminant is calculated as:

• \( b^2 - 4ac = 6^2 - 4(3)(-5) = 36 + 60 = 96 \)

A positive discriminant, like in this case, indicates there are two distinct real solutions. If the discriminant were zero, it would imply exactly one real solution, as the square root part of the formula would vanish, simplifying the expression. Conversely, if the discriminant were negative, the equation would have no real solutions, since you can't take the square root of a negative number within the real number system.
Understanding the discriminant is essential for predicting the type of solutions before even fully solving the quadratic equation.

Real solutions refer to the actual, tangible roots of a quadratic equation that can be plotted on the real number line. In this exercise, we've determined the quadratic equation \( 3x^2 + 6x - 5 = 0 \) has a positive discriminant (\( 96 \)), which means it has two distinct real solutions.
To find these solutions, we substitute back into the quadratic formula:

• \( x = \frac{-6 \pm \sqrt{96}}{6} \)

Calculating \( \sqrt{96} \) gives us \( 4\sqrt{6} \), leading to the solutions:

• \( x = \frac{-6 + 4\sqrt{6}}{6} = \frac{-3 + 2\sqrt{6}}{3} \)
• \( x = \frac{-6 - 4\sqrt{6}}{6} = \frac{-3 - 2\sqrt{6}}{3} \)

Both of these solutions are real and distinct, confirming our step-by-step assessment using the quadratic formula and the discriminant.