Problem 14 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ (t-2)^{2}=11 $$

Step-by-Step Solution

Answer

The solutions are \( t = 2 + \sqrt{11} \) and \( t = 2 - \sqrt{11} \).

Step 1: Expand the Left Side

The given quadratic equation is \((t-2)^2 = 11\). First, we recognize that the left side is a square of a binomial. We can expand this as \((t-2)(t-2)\), which yields: \[ t^2 - 4t + 4 \] Thus, the equation becomes \( t^2 - 4t + 4 = 11 \).

Step 2: Rearrange the Equation

Next, we move all terms to one side to set the equation to zero. Subtract 11 from both sides to get: \[ t^2 - 4t + 4 - 11 = 0 \] Simplifying further gives: \[ t^2 - 4t - 7 = 0 \].

Step 3: Use the Quadratic Formula

To solve the quadratic equation \( t^2 - 4t - 7 = 0 \), we use the quadratic formula, which is given by:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = -4 \), and \( c = -7 \). Substituting these values into the formula, we have:\[ t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-7)}}{2 \cdot 1} \].

Step 4: Simplify the Expression

Now calculate the expression under the square root, known as the discriminant:\[ (-4)^2 - 4 \cdot 1 \cdot (-7) = 16 + 28 = 44 \] Thus, the quadratic formula becomes:\[ t = \frac{4 \pm \sqrt{44}}{2} \].

Step 5: Simplify Further

Simplify \( \sqrt{44} \) to \( \sqrt{4 \times 11} = 2\sqrt{11} \). Substitute this back into the formula:\[ t = \frac{4 \pm 2\sqrt{11}}{2} \].Divide each term by 2 to get:\[ t = 2 \pm \sqrt{11} \].

Step 6: Check Your Solutions

The solutions are \( t = 2 + \sqrt{11} \) and \( t = 2 - \sqrt{11} \). To check these solutions, substitute them back into the original equation \((t-2)^2 = 11\):For \( t = 2 + \sqrt{11} \), \( (2 + \sqrt{11} - 2)^2 = (\sqrt{11})^2 = 11 \), which is correct.For \( t = 2 - \sqrt{11} \),\( (2 - \sqrt{11} - 2)^2 = (-\sqrt{11})^2 = 11 \), which is also correct. Both solutions satisfy the original equation.

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool in algebra that helps us find the solutions of quadratic equations in the form \( ax^2 + bx + c = 0 \). When solving for the variable \( x \), the quadratic formula is given by:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula provides two potential solutions, known as the roots of the equation, which are derived using the values of \( a \), \( b \), and \( c \). These components are coefficients in the quadratic equation, specifically the coefficients of \( x^2 \), \( x \), and the constant term, respectively.

The quadratic formula is especially useful when factoring the equation is difficult or impossible. To use the formula correctly, ensure that the equation is set to zero, as any discrepancies can lead to erroneous solutions. The formula works for all types of quadratic equations, whether they have two real roots, one real root, or complex roots.

Discriminant

The discriminant is part of the quadratic formula and plays a crucial role in determining the nature of the roots. It is the expression found under the square root in the quadratic formula:

• \( b^2 - 4ac \)

The discriminant provides insight into how many and what type of solutions the quadratic equation will have:

• If the discriminant is positive, there are two distinct real roots.
• If the discriminant is zero, there is exactly one real root, also known as a repeated or double root.
• If the discriminant is negative, the quadratic equation has two complex conjugate roots.

In the given exercise, we calculated the discriminant for the equation \( t^2 - 4t - 7 = 0 \) to be 44. Since 44 is positive, it confirms that there are two distinct real solutions for the equation.

Binomial Expansion

In the original exercise, the equation starts with a binomial square \((t-2)^2 = 11\). Recognizing this form allows us to use the binomial expansion, which involves expressing the binomial square as a quadratic. Specifically:

• \((a-b)^2 = a^2 - 2ab + b^2\)

For \((t-2)^2\), applying the binomial expansion yields:

• \(t^2 - 4t + 4\)

This step of expansion is vital because it transforms the equation into a standard quadratic form \( ax^2 + bx + c \) that we can manipulate with algebraic methods like the quadratic formula. Expanding binomials effectively sets the stage for rearranging the equation and applying subsequent solution methods.

Equation Solving Steps

Solving a quadratic equation generally follows a structured sequence of steps to ensure accuracy and efficiency:

• Identify the Equation Form: Ensure the equation is a quadratic and set to zero. In the exercise, it starts as \( (t-2)^2 - 11 = 0 \).
• Expand if Necessary: Expand any binomials. For example, \( (t-2)^2 \) becomes \( t^2 - 4t + 4 \).
• Rearrange the Equation: Move all terms to one side to form \( t^2 - 4t - 7 = 0 \).
• Apply the Quadratic Formula: Use the formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with the identified coefficients.
• Calculate the Discriminant: Compute \( b^2 - 4ac \) to understand the nature of the roots.
• Simplify the Solutions: Perform arithmetic to reduce terms and express the solutions in their simplest form.
• Verify the Solutions: Substitute the solutions back into the original equation to check validity.

Following these steps helps students systematically approach and solve quadratic equations, ensuring each aspect is addressed comprehensively.