Problem 52 - Intermediate Algebra

Question

For each quadratic equation, first use the discriminant to determine whether the equation has two nonreal complex solutions, one real solution with a multiplicity of two, or two real solutions. Then solve the equation. $$ x^{2}-3 x-54=0 $$

Step-by-Step Solution

Answer

The equation has two real solutions: \( x = 9 \) and \( x = -6 \).

Step 1: Write down the general form of a quadratic equation

The general form of a quadratic equation is \( ax^2 + bx + c = 0 \). For the given equation \( x^2 - 3x - 54 = 0 \), we identify \( a = 1 \), \( b = -3 \), and \( c = -54 \).

Step 2: Calculate the discriminant

The discriminant \( \Delta \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). Substitute the values of \( a \), \( b \), and \( c \) into this formula: \( \Delta = (-3)^2 - 4 \times 1 \times (-54) \).

Step 3: Evaluate the discriminant

Calculate \( \Delta = 9 + 216 = 225 \). The discriminant is positive, which means the quadratic equation has two distinct real solutions.

Step 4: Use the quadratic formula to find the roots

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = 1 \), \( b = -3 \), and \( \Delta = 225 \) into the formula: \( x = \frac{-(-3) \pm \sqrt{225}}{2 \times 1} \).

Step 5: Calculate the solutions

Calculate \( x = \frac{3 \pm 15}{2} \). This gives two solutions: \( x_1 = \frac{3 + 15}{2} = 9 \) and \( x_2 = \frac{3 - 15}{2} = -6 \).

Key Concepts

Discriminant

In quadratic equations, the discriminant plays a crucial role in predicting the nature of its roots. It is a key component derived from the equation. The discriminant is calculated as \( \Delta = b^2 - 4ac \) from the standard form of a quadratic equation \( ax^2 + bx + c = 0 \). Depending on the value of the discriminant, we can determine the type of solutions the quadratic equation will have.
If \( \Delta > 0 \), the quadratic equation has two distinct real solutions, meaning the parabola intersects the x-axis at two points. If \( \Delta = 0 \), there is exactly one real solution with multiplicity two, meaning the parabola touches the x-axis at just one point. Lastly, if \( \Delta < 0 \), the quadratic equation has two complex non-real solutions, signifying that the parabola does not intersect the x-axis at all.
In the original exercise, the discriminant was calculated as 225, indicating two distinct real solutions because 225 is a positive number.

Quadratic Formula

The quadratic formula is like a magical tool for solving all sorts of quadratic equations. Once you've found the discriminant, you can use the quadratic formula to find the exact solutions, or roots, of the equation. The formula is expressed as:

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

Substituting the values of \( a \), \( b \), and \( \Delta \) (or \( b^2 - 4ac \)) into this formula will allow you to find the values of \( x \) that make the quadratic equation true.
In the exercise, \( a = 1 \), \( b = -3 \), and \( \Delta = 225 \). By plugging these numbers into the quadratic formula, you have \( x = \frac{3 \pm 15}{2} \). This will lead to two solutions: \( x_1 = 9 \) and \( x_2 = -6 \). These solutions represent the x-coordinates where the parabola described by the quadratic equation crosses the x-axis.

Real Solutions

Real solutions refer to the values of \( x \) that satisfy the quadratic equation, resulting in real numbers. In the context of quadratic equations, having real solutions is significant because it determines how the graph of the equation behaves on the coordinate plane.
When we talk about real solutions, we are specifically referring to the x-values where the parabola touches or intersects the x-axis. This occurs when the discriminant is non-negative (\( \Delta \geq 0 \)).

• Two real solutions: If the discriminant is greater than zero (\( \Delta > 0 \)), the quadratic equation has two distinct real numbers as solutions, indicating the parabola intersects the x-axis at two different points.
• One real solution: If the discriminant equals zero (\( \Delta = 0 \)), there is one unique real solution, which means the parabola just touches the x-axis without crossing it.

In the given exercise, the quadratic equation \( x^2 - 3x - 54 = 0 \) has a positive discriminant, leading to two real solutions: \( x = 9 \) and \( x = -6 \). These values are the points where the parabola intersects the x-axis.