Problem 2 - Algebra for College Students

Question

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 distinct real solutions: \(x = 9\) and \(x = -6\).

Step 1: Identify the Coefficients

Given the quadratic equation \(x^2 - 3x - 54 = 0\), we can identify the coefficients as follows: \(a = 1\), \(b = -3\), and \(c = -54\). These are crucial for computing the discriminant.

Step 2: Calculate the Discriminant

The discriminant \(\Delta\) of a quadratic equation \(ax^2 + bx + c = 0\) is given by the formula \(\Delta = b^2 - 4ac\). Substitute the values of \(a\), \(b\), and \(c\): \(\Delta = (-3)^2 - 4 \cdot 1 \cdot (-54) = 9 + 216 = 225\).

Step 3: Interpret the Discriminant

The value of the discriminant \(\Delta = 225\) is greater than zero, indicating that the quadratic equation has two distinct real solutions.

Step 4: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). We have already calculated \(b^2 - 4ac = 225\), so substitute into the formula: \(x = \frac{-(-3) \pm \sqrt{225}}{2 \times 1}\). This simplifies to \(x = \frac{3 \pm 15}{2}\).

Step 5: Calculate the Solutions

Using the quadratic formula solutions, we have two possible solutions:1. \(x = \frac{3 + 15}{2} = \frac{18}{2} = 9\)2. \(x = \frac{3 - 15}{2} = \frac{-12}{2} = -6\).

Step 6: Verify the Solutions

Substitute the solutions back into the original equation:- For \(x = 9\): \(9^2 - 3 \times 9 - 54 = 81 - 27 - 54 = 0\)- For \(x = -6\): \((-6)^2 - 3 \times (-6) - 54 = 36 + 18 - 54 = 0\). Both satisfy the equality, confirming the solutions.

Key Concepts

Understanding the Discriminant

When solving quadratic equations of the form \( ax^2 + bx + c = 0 \), the discriminant provides key insights. It is found inside the square root of the quadratic formula as \( b^2 - 4ac \). This expression, called the discriminant, helps determine the nature of the roots:

• If the discriminant is positive (\( \Delta > 0 \)), the equation has two distinct real solutions.
• If it equals zero (\( \Delta = 0 \)), there is exactly one real solution, also known as a repeated or double root.
• If negative (\( \Delta < 0 \)), the solutions are complex or non-real.

For instance, in the equation \( x^2 - 3x - 54 = 0 \), the discriminant was calculated to be 225. Since this is a positive number, we know the quadratic has two real solutions.

Utilizing the Quadratic Formula

The quadratic formula is your go-to tool for finding solutions to any quadratic equation. It goes as follows: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula uses the discriminant to decide if the solutions are real and how many there are.
In our example equation \( x^2 - 3x - 54 = 0 \), we plug the coefficients \( a = 1 \), \( b = -3 \), \( c = -54 \) into the formula: \[ x = \frac{-(-3) \pm \sqrt{225}}{2 \times 1} \]Simplifying further, this becomes: \[ x = \frac{3 \pm 15}{2} \]From there, you find the possible solutions:

• First solution: \( \frac{3 + 15}{2} = 9 \)
• Second solution: \( \frac{3 - 15}{2} = -6 \)

Thus, the quadratic formula not only pinpoints our solutions but confirms their real nature.

Identifying Real Solutions

Real solutions mean the roots of the equation are numbers you can plot on a number line, as opposed to complex ones which include imaginary numbers. When the discriminant is positive, both solutions from the quadratic formula will be real numbers.
This is evident in the steps of our equation, \( x^2 - 3x - 54 = 0 \), where the discriminant, \( 225 \), signified two real roots. Solving it provided solutions \( x = 9 \) and \( x = -6 \).
To confirm these as valid solutions, we substitute each solution back into the original equation to check if it satisfies the equation:

• For \( x = 9 \), substituting returns \( 81 - 27 - 54 = 0 \).
• For \( x = -6 \), substituting returns \( 36 + 18 - 54 = 0 \).

Both result in true statements, verifying they are indeed the correct real solutions. Understanding and verifying real solutions is crucial as it ensures the integrity of our solutions and bolsters our algebraic solving skills.