Problem 14 - Basic Technical Mathematics with Calculus

Question

Solve the given quadratic equations by completing the square. Exercises \(11-14\) and \(17-20\) may be checked by factoring. $$t^{2}+5 t-6=0$$

Step-by-Step Solution

Answer

The solutions are \(t = 1\) and \(t = -6\).

Step 1: Move Constant to Right Side

Start with the equation \(t^2 + 5t - 6 = 0\). Move the constant term \(-6\) to the right side of the equation to focus on the quadratic and linear terms. This gives us: \(t^2 + 5t = 6\).

Step 2: Prepare to Complete the Square

Take the coefficient of \(t\), which is \(5\), divide it by \(2\) to get \(\frac{5}{2}\), and then square it to form a perfect square trinomial. This results in \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\).

Step 3: Adjust Both Sides

Add \(\frac{25}{4}\) to both sides of the equation to maintain balance: \(t^2 + 5t + \frac{25}{4} = 6 + \frac{25}{4}\). Simplify the right side: \(6\) is \(\frac{24}{4}\), so \(6 + \frac{25}{4} = \frac{49}{4}\). Now, we have: \(t^2 + 5t + \frac{25}{4} = \frac{49}{4}\).

Step 4: Factor Left Side

The left side \(t^2 + 5t + \frac{25}{4}\) is a perfect square trinomial and can be factored into \((t + \frac{5}{2})^2\). Thus, the equation becomes: \((t + \frac{5}{2})^2 = \frac{49}{4}\).

Step 5: Solve for \(t\) Using Square Roots

To isolate \(t\), take the square root of both sides. Note that there are two solutions when taking the square root: \[t + \frac{5}{2} = \frac{7}{2} \quad \text{or} \quad t + \frac{5}{2} = -\frac{7}{2}\].

Step 6: Solve Equations for \(t\)

Solve the equations \(t + \frac{5}{2} = \frac{7}{2}\) and \(t + \frac{5}{2} = -\frac{7}{2}\). For the first equation: subtract \(\frac{5}{2}\) from both sides to get \(t = \frac{7}{2} - \frac{5}{2} = 1\). For the second equation: \(t = -\frac{7}{2} - \frac{5}{2} = -6\).

Step 7: Conclusion: Verify Solutions

The solutions to the quadratic equation are \(t = 1\) and \(t = -6\). These solutions can be verified by checking that they satisfy the original equation \(t^2 + 5t - 6 = 0\). Substituting \(t = 1\) results in \((1)^2 + 5(1) - 6 = 0\) and \(t = -6\) results in \((-6)^2 + 5(-6) - 6 = 0\), confirming both are correct.

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra. They are polynomial equations of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). In these equations:

• "\( ax^2 \)" is the quadratic term.
• "\( bx \)" is the linear term.
• "\( c \)" is the constant term.

Quadratic equations can have up to two real solutions. Solving these equations can involve several methods, including factoring, completing the square, or using the quadratic formula.
Understanding quadratic equations is crucial because they enable us to model and solve various real-world problems, such as trajectory paths, areas, and optimization problems.

Factoring

Factoring is a technique used to simplify quadratic equations. It involves expressing the quadratic equation as a product of two binomial expressions. If the quadratic equation can be written as \((px + q)(rx + s) = 0\), solving it becomes straightforward. Here’s how it works:

• Find two numbers that multiply to give \(a \cdot c\) and add up to \(b\).
• Rewrite the middle term using these two numbers.
• Factor by grouping and simplify.

Factoring is effective when dealing with simple numbers that lead to integer binomial factors. However, not all quadratic equations are easily factorable, which is when completing the square or other methods are helpful.
Like in the original exercise, once we have the equation in a form that can be factored, we can solve for the variable by setting each factor equal to zero and solving those simpler equations.

Perfect Square Trinomial

A perfect square trinomial is a special quadratic expression that can be written as the square of a binomial. It takes the form \((x + d)^2 = x^2 + 2dx + d^2\). Recognizing a perfect square trinomial allows for easier manipulation and solving of equations. Here's the process:

• Take the coefficient of the linear term, divide it by 2.
• Square that result to find the constant needed to complete the square.

For example, in the equation \(t^2 + 5t\), the coefficient of \(t\) is 5. Dividing by 2 gives \(\frac{5}{2}\), and squaring it results in \(\frac{25}{4}\). Adding this to both sides transforms the expression into a perfect square trinomial.
Once identified or created, it simplifies the process of solving a quadratic equation by creating a base to take the square root of both sides.

Solving Equations

Solving quadratic equations is about finding the values of the variable that make the equation true. When we solve equations by completing the square, we transform the equation into a perfect square trinomial. This method is a valuable tool when factoring is challenging or impossible.
Here’s a step-by-step look at this method:

• Move the constant term to the opposite side of the equation.
• Complete the square on the variable terms (as explained in the perfect square trinomial section).
• Transform the equation into a perfect square trinomial and take the square root of both sides.
• Solve for the variable using the new linear equations derived from each side of the equation.

Remember, when you solve quadratic equations and take square roots, consider both positive and negative solutions, since both can satisfy the original equation.
This method helps ensure that no potential solutions are overlooked and gives a complete solution set for the equation.