Problem 31 - Elementary and Intermediate Algebra

Question

Solve each equation. $$ x^{2}-9 x+8=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 8 \) and \( x = 1 \).

Step 1: Identify the Quadratic Formula

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \). Here, the coefficients are \( a=1 \), \( b=-9 \), and \( c=8 \). We use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] to find the solutions for \( x \).

Step 2: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \). \[ \text{Discriminant} = (-9)^2 - 4 \times 1 \times 8 = 81 - 32 = 49 \] Since the discriminant is positive, there will be two real and distinct solutions.

Step 3: Apply the Quadratic Formula

Substitute the values of \( a \), \( b \), and \( c \) into the quadratic formula. \[ x = \frac{-(-9) \pm \sqrt{49}}{2 \times 1} \] Simplifies to: \[ x = \frac{9 \pm 7}{2} \]

Step 4: Solve for the Roots

Now solve for the two roots using the \( \pm \) sign: First root: \[ x = \frac{9 + 7}{2} = \frac{16}{2} = 8 \] Second root: \[ x = \frac{9 - 7}{2} = \frac{2}{2} = 1 \] Thus, the two solutions are \( x = 8 \) and \( x = 1 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). A quadratic equation is one where the highest exponent of the variable \( x \) is 2. This formula provides a straightforward way to find the roots or solutions of the equation. The quadratic formula is given as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula helps us find the values of \( x \) that make the equation true. It works by using the coefficients \( a \), \( b \), and \( c \) from the quadratic equation. Here's a brief overview of how the components work:

• \( a \) is the coefficient of \( x^2 \)
• \( b \) is the coefficient of \( x \)
• \( c \) is the constant term

The plus-minus symbol (\( \pm \)) in the formula means there are generally two solutions to the equation because you have to calculate it once with a plus and once with a minus. This is where the concept of the discriminant comes into play, which helps determine the nature of these solutions.

Discriminant

The discriminant is an essential part of solving quadratic equations using the quadratic formula. It's found under the square root symbol in the formula and is represented as:\[b^2 - 4ac\]The value of the discriminant reveals the nature of the roots of the quadratic equation. Here's how:

• If the discriminant is positive, \( (b^2 - 4ac > 0) \), there are two real and distinct solutions. This means the quadratic equation crosses the x-axis at two different points.
• If the discriminant is zero, \( (b^2 - 4ac = 0) \), there is exactly one real solution, also known as a repeated or double root. The graph of the equation touches the x-axis at exactly one point.
• If the discriminant is negative, \( (b^2 - 4ac < 0) \), there are no real solutions, and the roots are complex or imaginary. In this case, the graph does not intersect the x-axis.

Understanding the value of the discriminant simplifies the process of predicting the type of solutions before even solving the equation.

Real and Distinct Solutions

When we talk about real and distinct solutions in the context of quadratic equations, we're referring to scenarios where the solutions are two different, real numbers.In the quadratic formula, the presence of a positive discriminant \( (b^2 - 4ac > 0) \) indicates two distinct solutions. These solutions are found by evaluating the expression twice: once with a plus sign and once with a minus sign.For the quadratic equation \( x^2 - 9x + 8 = 0 \), the discriminant is 49, which is positive. Therefore, the roots are real and distinct. By substituting the values into the quadratic formula, we get:

• First solution: \( x = \frac{9 + 7}{2} = 8 \)
• Second solution: \( x = \frac{9 - 7}{2} = 1 \)

These solutions refer to the points where the curve of the quadratic equation intersects the x-axis. Real and distinct solutions indicate that the parabola opens outward, creating two intersections on the x-axis, confirming that these solutions are not only real but also different from each other.