Problem 51 - Precalculus : Building Concepts and Connections

Question

Solve the quadratic equation by using the quadratic formula. Find only real solutions. $$\frac{1}{2} t^{2}-4 t-3=0$$

Step-by-Step Solution

Answer

The real solutions for the given quadratic equation are \( t = 8.69 \) and \( t = -0.69 \)

Step 1: Identify the Coefficients

For the quadratic equation \( \frac{1}{2} t^{2}-4 t-3=0 \), the coefficients are \( a= \frac{1}{2} \), \( b= -4 \), and \( c= -3 \).

Step 2: Calculate the Discriminant

The discriminant is given by the formula \( D=b^{2}-4ac \). Substituting the values for \( a \), \( b \), and \( c \), we get: \( D=(-4)^{2}-4* \( \frac{1}{2} \)*(-3) = 16 + 6 = 22\).

Step 3: Apply the Quadratic Formula

The quadratic formula is \( t= \frac{-b± \sqrt{D}}{2a} \). Given that D is 22, and \( a= \frac{1}{2}, b=-4 \), we can substitute these values.

Step 4: Calculate the Real Solutions

Using substitutions, we have two possible solutions for \( t \): \( t= \frac{-(-4) + \sqrt{22}}{2* \( \frac{1}{2} \)} = 4+ \sqrt{22} = 4 \pm 4.694 \). Therefore, the real solutions for \( t \) are \( 8.69 \) and \( -0.69 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula provides a way to find the values of \( x \) that satisfy the equation. The quadratic formula is represented as:
\[\[\begin{equation}x = \frac{-b \pm \sqrt{D}}{2a}\end{equation}\]\]Here, \( a \), \( b \), and \( c \) are coefficients from the quadratic equation. The symbol \( \pm \) means "plus or minus," which indicates that there will typically be two solutions. Using the quadratic formula helps solve the equation by calculating these two potential values of \( x \) where the equation equals zero.
This method is especially useful when factoring is difficult or when the coefficients are inconvenient. Always remember to correctly identify \( a \), \( b \), and \( c \) from the equation before applying the formula.

Real Solutions

Real solutions of quadratic equations are solutions that are real numbers, as opposed to imaginary numbers. They can be obtained using the quadratic formula. Whether these solutions are real or imaginary depends on the discriminant (\( D \)) that appears under the square root in the quadratic formula:
\[\sqrt{D}\]When the discriminant \( D \) is greater than zero, the quadratic equation has two distinct real solutions. If \( D \) is equal to zero, there is one real solution, which implies that the solutions are repeated, or "double solutions." When \( D \) is less than zero, the solutions are not real but rather imaginary, which means they involve the square root of a negative number.
Understanding real solutions helps in analyzing the nature of the quadratic equation and its solutions when applying the quadratic formula.

Discriminant Calculation

The discriminant is a key part of understanding quadratic equations and their solutions. It affects how many and what type of solutions an equation has. The discriminant \( D \) is calculated using the formula:
\[D = b^2 - 4ac\]In this formula, \( a \), \( b \), and \( c \) refer to the coefficients of the quadratic equation. Calculating \( D \) helps predict the nature of the solutions:

• If \( D > 0 \), the quadratic equation has two distinct real solutions.
• If \( D = 0 \), there is exactly one real solution, resulting in a double root where the parabola touches the x-axis.
• If \( D < 0 \), the solutions are imaginary, indicating that the parabola does not intersect the x-axis.

Computing the discriminant is the first step to determining how the quadratic formula should be applied in order to find the solutions to a given quadratic equation.