Problem 41 - Beginning and Intermediate Algebra

Question

Solve each equation. $$b(b-4)=96$$

Step-by-Step Solution

Answer

The two solutions for \(b\) in the equation \(b^2 - 4b - 96 = 0\) are \(b = 12\) and \(b = -8\).

Step 1: Simplify the equation

Multiply out the brackets: \(b(b-4) = 96\) \(b^2 - 4b - 96 = 0\) Now, our equation is in the standard quadratic form: \(b^2 - 4b - 96=0\).

Step 2: Determine if it's factorable or use Quadratic Formula

Check if the quadratic equation can be factored easily. If not, use the quadratic formula \(b = \frac{-c \pm \sqrt{c^2 - 4ad}}{2a}\). We can't find two numbers that multiply to -96 and add to -4 quickly, so we'll use the quadratic formula with \(a = 1\), \(c = -4\), and \(d = -96\): \(b = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-96)}}{2(1)}\)

Step 3: Solve for b using the Quadratic Formula

Plug in the values for a, c, and d into the quadratic formula. \(b = \frac{4 \pm \sqrt{16 + 384}}{2}\)

Step 4: Simplify the expression under the square root

Add the numbers inside the square root: \(b = \frac{4 \pm \sqrt{400}}{2}\)

Step 5: Simplify the square root

Take the square root of the number inside the square root: \(b = \frac{4 \pm 20}{2}\) We have two possible solutions for b:

Step 6: Calculate both solutions

Calculate the two possible values for b: \(b_1 = \frac{4 + 20}{2} = \frac{24}{2} = 12\) \(b_2 = \frac{4 - 20}{2} = \frac{-16}{2} = -8\) The two solutions for \(b\) are \(b = 12\) and \(b = -8\).

Key Concepts

Quadratic Formula

The quadratic formula is one of the most powerful tools for solving quadratic equations. Quadratic equations typically take the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients, and \( x \) represents the variable whose values we are solving for. The quadratic formula allows us to find the solutions (also known as roots) of any quadratic equation. It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]This formula might look a bit intimidating at first, but its structure helps to determine the roots in a straightforward way. The key part of the formula is the discriminant, \( b^2 - 4ac \). This part determines the nature and number of the roots:

• If \( b^2 - 4ac > 0 \), there are two distinct real roots.
• If \( b^2 - 4ac = 0 \), there is exactly one real root, also known as a repeated or double root.
• If \( b^2 - 4ac < 0 \), the roots are complex (not real).

In the original exercise, we used the quadratic formula because factoring wasn't straightforward, demonstrating its utility in providing exact solutions under any circumstance.

Factoring Quadratics

Factoring quadratics is a method to solve quadratic equations by rewriting them in a product form. A quadratic equation can often be expressed as a multiplication of two binomials. For example, an equation \( ax^2 + bx + c = 0 \) can sometimes be rewritten as \((mx + n)(px + q) = 0\). The key is to find numbers \( m, n, p, \) and \( q \) that satisfy:

• \( m \times p = a \)
• \( n \times q = c \)
• \( m \times q + n \times p = b \)

In practice, certain quadratics can easily be factored by inspection, especially when the coefficients are simple integers. However, in our exercise \( b^2 - 4b - 96 = 0 \), finding two numbers that achieve this is not immediate, which led us to the quadratic formula approach. That said, factoring is useful when applicable, especially for straightforward quadratic expressions. It offers a quicker route to the roots if the equation can be factored.

Standard Quadratic Form

The standard quadratic form is essential for identifying and solving quadratic equations. It is represented as \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants, and 'x' is the variable term.Putting an equation into this form consists of rearranging all terms so that they are on one side of the equal sign, with zero on the other. In our example, the equation started in the form \( b(b-4) = 96 \). After expanding and moving all terms to one side, we obtained the standard quadratic form: \( b^2 - 4b - 96 = 0 \).This form is incredibly useful because it lays the groundwork for using methods such as factoring, completing the square, or applying the quadratic formula to find the roots of the equation. Once in this form, the equation is ready for further analysis and solution methods.

Roots of Quadratic Equation

The roots of a quadratic equation are the solutions where the equation equals zero. These values of \( x \) make the quadratic expression \( ax^2 + bx + c = 0 \) true. Finding these roots or solutions is fundamental as they indicate the points where the graph of the quadratic function intersects the x-axis.As mentioned before, the nature of the roots depends on the discriminant \( b^2 - 4ac \) of the quadratic equation:

• Two distinct real roots if \( b^2 - 4ac > 0 \)
• One repeated real root if \( b^2 - 4ac = 0 \)
• Complex roots if \( b^2 - 4ac < 0 \)

In our exercise, the quadratic equation \( b^2 - 4b - 96 = 0 \) had two distinct real roots: \( b = 12 \) and \( b = -8 \). This means the parabola corresponding to this equation crosses the x-axis at these two points, representing the solutions to the equation.