Problem 42 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ 2 y^{2}-y-\frac{1}{2}=0 $$

Step-by-Step Solution

Answer

The real solutions are \( y = \frac{1 + \sqrt{5}}{4} \) and \( y = \frac{1 - \sqrt{5}}{4} \).

Step 1: Identify Coefficients

Start by identifying the coefficients in the quadratic equation of the form \( ax^2 + bx + c = 0 \). For the given equation \( 2y^2 - y - \frac{1}{2} = 0 \), the coefficients are \( a = 2 \), \( b = -1 \), and \( c = -\frac{1}{2} \).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( b^2 - 4ac \). Substituting the values here: \( b^2 - 4ac = (-1)^2 - 4 \times 2 \times \left(-\frac{1}{2}\right) \). Calculate this to get \( 1 + 4 = 5 \).

Step 3: Apply the Quadratic Formula

The solutions of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substituting the values of \( a \), \( b \), and the discriminant: \( y = \frac{-(-1) \pm \sqrt{5}}{2 \times 2} \). Simplify to get \( y = \frac{1 \pm \sqrt{5}}{4} \).

Step 4: Find the Real Solutions

Simplify the expressions for the real solutions using the solutions found from the quadratic formula. The solutions are \( y = \frac{1 + \sqrt{5}}{4} \) and \( y = \frac{1 - \sqrt{5}}{4} \).

Key Concepts

Real Solutions

Real solutions of a quadratic equation are the values of the variable that satisfy the equation, found using methods like factoring, completing the square, or using the quadratic formula. In mathematics, a solution is called "real" if it belongs to the set of real numbers, meaning it can be plotted on a number line.
In some cases, a quadratic equation may not have real solutions, and instead, it has complex or imaginary solutions. However, when the discriminant (explained later) is positive or zero, we can be certain that the solutions are real.
In our problem, after applying the quadratic formula, we found two real solutions for the equation. These solutions are:

• \( y = \frac{1 + \sqrt{5}}{4} \)
• \( y = \frac{1 - \sqrt{5}}{4} \)

Both of these numbers can be shown on a real number line, confirming they are indeed real solutions.

Quadratic Formula

The quadratic formula is a powerful tool that provides a straightforward method to solve any quadratic equation. Given a quadratic equation in the form \( ax^2 + bx + c = 0 \), the quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula helps find the solutions, or roots, of any quadratic equation, whether the solutions are real or complex.
To effectively use the quadratic formula, we need to identify the coefficients of the equation: \( a \), \( b \), and \( c \). Once these coefficients are substituted into the equation, and the square root of the discriminant is calculated, the formula directly gives the two possible solutions.
In our example with \( 2 y^{2} - y - \frac{1}{2} = 0 \), using the formula resulted in two solutions, demonstrating the utility of this approach when dealing with quadratic equations.

Discriminant

The discriminant, denoted as \( \Delta \), plays a critical role in determining the nature of the solutions of a quadratic equation. The discriminant is given by the formula \( b^2 - 4ac \). It is part of the quadratic formula and influences the ultimate nature of the solutions.

• If \( \Delta > 0 \), there are two distinct and real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (or a repeated root).
• If \( \Delta < 0 \), there are no real solutions, only complex ones.

In our specific problem, \( \Delta = 5 \), which is greater than zero. This indicates that there are two distinct real solutions. Recognizing the sign and value of the discriminant helps predict the type of solutions before even applying the quadratic formula.