Problem 43 - Beginning and Intermediate Algebra

Question

Solve each equation. $$-63=4 j(j-8)$$

Step-by-Step Solution

Answer

The solutions for the equation \(-63 = 4j(j - 8)\) are \(j = 4.5\) and \(j = 3.5\).

Step 1: Expand the Equation

To start, let's expand the given equation, \(-63 = 4j(j - 8)\): \[ -63 = 4j^2 - 32j \]

Step 2: Transform into Standard Quadratic Form

Now, let's rearrange the equation into a standard quadratic equation form, \(ax^2 + bx + c = 0\): \[ 4j^2 - 32j + 63 = 0 \]

Step 3: Solving the Quadratic Equation using Quadratic Formula

To solve for j, we can use the quadratic formula \(\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 4\), \(b = -32\), and \(c = 63\). Substitute the values of a, b, and c into the quadratic formula: \[j = \frac{-(-32) \pm \sqrt{(-32)^2 - 4(4)(63)}}{2(4)}\]

Step 4: Simplify the Quadratic Formula

Now we can simplify the quadratic formula to get our values for j: \[j = \frac{32 \pm \sqrt{1024 - 1008}}{8}\] \[j = \frac{32 \pm \sqrt{16}}{8}\]

Step 5: Determine Possible Values for j

Now we can solve for the two possible values of j: \[j = \frac{32 + 4}{8} = \frac{36}{8} = 4.5\] \[j = \frac{32 - 4}{8} = \frac{28}{8} = 3.5\]

Step 6: Conclusion

The solutions for the equation \(-63 = 4j(j - 8)\) are: \[j = 4.5\] and \[j = 3.5\]

Key Concepts

Solving Quadratic Equations

Quadratic equations are essential in algebra and involve expressions that follow the standard form: \(ax^2 + bx + c = 0\). These equations represent parabolas when graphed. Solving them involves finding values of \(x\) (or any other variable) that satisfy this equation.

The equation \(-63 = 4j(j-8)\) initially might look intimidating. However, by systematically applying the principles of solving quadratic equations, we can find a clear path to the solution.

• First, make sure the equation is expanded and simplified, bringing all terms to one side, thus setting up for the standard form.
• Next, apply formulas like factoring, completing the square, or using the quadratic formula to find solutions.

Once the equation is arranged correctly, the focus shifts to finding the correct values for \(j\) that make the equation true.

Quadratic Formula

The quadratic formula is a powerful tool for solving equations when factoring is too complex or impossible. The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This renowned formula calculates the roots of any quadratic equation by substituting values \(a\), \(b\), and \(c\) from the equation. In the example \(4j^2 - 32j + 63 = 0\), those values are \(a = 4\), \(b = -32\), and \(c = 63\).

Here's a breakdown of how it works:

• Calculate the discriminant, \(b^2 - 4ac\). This tells us how many solutions exist: a positive discriminant indicates two solutions, zero a single solution, and a negative none.
• Compute the solutions using the full formula, considering both the positive and negative outcomes from the square root.

The quadratic formula is critical in algebra, providing exact solutions efficiently.

Expanding Equations

Expanding equations is often a necessary step in transforming a complex equation into a familiar form. It involves distributing and simplifying terms to reveal a pattern that can be solved more effectively.

In our example, \(-63=4j(j-8)\), expanding was the first task:

• Multiply each term inside the parentheses by \(4j\): \(4j \cdot j\) and \(4j \cdot -8\).
• This results in \(-63 = 4j^2 - 32j\), a form much easier to manage.

Expanding not only clarifies the structure of the equation but allows us to apply methods like completing the square or using the quadratic formula more effectively.

Standard Quadratic Form

Any quadratic equation can be expressed in a standard form: \(ax^2 + bx + c = 0\). This format helps identify what's required for solution methods like factoring or using the quadratic formula.

For the equation \(-63 = 4j(j - 8)\), transitioning to the standard form involved bringing all terms to one side:

• Move \(-63\) across the equals sign, resulting in \(4j^2 - 32j + 63 = 0\).

This transformation is pivotal. Once in standard form, the coefficients \(a\), \(b\), and \(c\) can be easily extracted, allowing for immediate advanced solution strategies.