Problem 47 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ \left(\frac{t}{t+1}\right)^{2}-\frac{2 t}{t+1}-8=0 $$

Step-by-Step Solution

Answer

The solutions are \( t = -\frac{4}{3} \) and \( t = -\frac{2}{3} \).

Step 1: Identify the Equation Type

The equation \( \left(\frac{t}{t+1}\right)^{2}-\frac{2 t}{t+1}-8=0 \) can be seen to be quadratic in nature with respect to the expression \(\frac{t}{t+1}\). Let's set \( x = \frac{t}{t+1} \). Then the equation becomes \( x^{2} - 2x - 8 = 0 \).

Step 2: Solve the Quadratic Equation

Now solve the quadratic equation \( x^{2} - 2x - 8 = 0 \). We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \), where \( a = 1 \), \( b = -2 \), and \( c = -8 \). Calculate the discriminant: \( b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-8) = 4 + 32 = 36 \). Therefore, \( x = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2} \). This gives \( x = 4 \) and \( x = -2 \).

Step 3: Substitute Back and Solve for t

We found values of \( x \), which correspond to \( \frac{t}{t+1} \). Let's substitute back.- For \( x = 4 \): \( \frac{t}{t+1} = 4 \), multiply both sides by \( t+1 \): \( t = 4t + 4 \), solving for \( t \) gives \( -3t = 4 \), so \( t = -\frac{4}{3} \).- For \( x = -2 \): \( \frac{t}{t+1} = -2 \), multiply both sides by \( t+1 \): \( t = -2t - 2 \), solving for \( t \) gives \( 3t = -2 \), so \( t = -\frac{2}{3} \).

Step 4: Verify Solutions

Plug \( t = -\frac{4}{3} \) and \( t = -\frac{2}{3} \) back into the original equation to verify the solutions:- For \( t = -\frac{4}{3} \): \( \left( \frac{-\frac{4}{3}}{-\frac{4}{3}+1} \right)^2 - \frac{2(-\frac{4}{3})}{-\frac{4}{3}+1} - 8 = \left( 4 \right) - (8) - 8 = 0 \).- For \( t = -\frac{2}{3} \): \( \left( \frac{-\frac{2}{3}}{-\frac{2}{3}+1} \right)^2 - \frac{2(-\frac{2}{3})}{-\frac{2}{3}+1} - 8 = \left( -2 \right) + (4) - 8 = 0 \).Both values satisfy the original equation.

Key Concepts

Fractional Equations

Fractional equations involve fractions that include variables in the numerator, the denominator, or both. In such equations, it's important to handle the denominators carefully to avoid undefined expressions.
In the given problem, the equation involves the fraction \( \frac{t}{t+1} \). The key to solving it lies in recognizing it as a quadratic equation, but first, we need to manage the fractional part.
When working with fractional equations, always consider possible restrictions based on the denominator. Ensure the denominator never equals zero, hence the expression \( t+1 eq 0 \) implies \( t eq -1 \).
Keep this restriction in mind, as it defines the values that \( t \) cannot take. This critical concept ensures that when solving the equation, you don't end up with invalid solutions.

Step-by-Step Solutions

Solving algebraic equations step-by-step ensures a clear understanding and makes complex problems manageable. Each solution step follows logically from the previous one until we reach the final answer.
Let's walk through the provided problem:

• Recognize the Equation Type: Identify the equation as quadratic by focusing on the expression \( \left(\frac{t}{t+1}\right)^2 - \frac{2t}{t+1} - 8 = 0 \). It helps simplify by letting \( x = \frac{t}{t+1} \). Now, you have \( x^2 - 2x - 8 = 0 \).
• Use the Quadratic Formula: Solve \( x^2 - 2x - 8 = 0 \) using the quadratic formula. Calculate the discriminant \( (b^2 - 4ac) \) to determine the nature of the roots.
• Find the Roots: Calculate roots \( x = 4 \) and \( x = -2 \), ensuring you compute accurately.
• Substitution and Verifying: Substitute back the roots into the original variables and verify by plugging them back into the equation.

This structured approach provides clarity and reduces errors.

Critical Points in Algebra

Understanding critical points in algebra is vital as they often signify important changes or boundaries within equations. These are particularly useful in quadratic equations where roots signify where the graph of the equation crosses the x-axis.
In the context of our fractional equation, setting \( x = 4 \) and \( x = -2 \) amends the expression to a solvable form. These critical points guide you in finding valid solutions for \( t \).
Once the values for \( x \) are determined, they serve as a pivot for substitution back into the original fraction-based expression. Thus, critical points provide direction during problem-solving, ensuring you arrive at valid solutions.

Substitution Method in Algebra

The substitution method is a powerful technique used to simplify and solve equations. It involves replacing a complex expression with a simpler variable to manage calculations easily.
For the problem at hand, the substitution \( x = \frac{t}{t+1} \) simplifies the equation into a standard quadratic form, \( x^2 - 2x - 8 = 0 \). This step converts a potentially complicated problem into a familiar territory.
Once you solve the simplified equation, reverse the substitution to find the values of the original variables. By setting \( \frac{t}{t+1} = 4 \) and \( \frac{t}{t+1} = -2 \), we return to the variable \( t \). Substitute back and solve to find \( t = -\frac{4}{3} \) and \( t = -\frac{2}{3} \).
This method reduces complexity and is essential for solving equations involving fractions or nested expressions.