Problem 66 - Algebra Form and Function

Question

In Problems \(59-70,\) decide for what values of the constant \(A\) the equation has (a) The solution \(t=0\) (b) A positive solution (c) A negative solution $$ t^{3}=-A^{2} $$

Step-by-Step Solution

Answer

Question: For the cubic equation \(t^3 = -A^2\), find the value of A for which the equation has (a) \(t=0\), (b) a positive solution, and (c) a negative solution. Answer: (a) A=0, (b) None, (c) A = \(\pm \sqrt{-t_n^3}\) for every \(t_n < 0\).

Step 1: Determine the value of A for t=0

Substitute \(t=0\) in the equation: $$ 0^3 = -A^2 \Rightarrow -A^2 = 0. $$ To find the value/s of \(A\), simply solve for \(A\): $$ A^2 = 0 \Rightarrow A = 0. $$ So, the equation has \(t=0\) when \(A=0\).

Step 2: Determine condition(s) of A for a positive solution

Suppose \(t = t_p>0\). We want to find values of \(A\) such that \(t^3 = -A^2\) has a positive solution. In other words, we want to find values of \(A\) that satisfy: $$ t_p^3 = -A^2. $$ Since \(t_p > 0\), then \(t_p^3 > 0\) also. However, \(-A^2\) is always non-positive, meaning we cannot have a positive solution \(t_p\) with this equation. Thus, there is no value of \(A\) for which the equation has a positive solution.

Step 3: Determine condition(s) of A for a negative solution

Suppose \(t = t_n<0\). We want to find values of \(A\) such that \(t^3 = -A^2\) has a negative solution. So, we aim to find values of \(A\) that satisfy: $$ t_n^3 = -A^2. $$ Since \(t_n < 0\), then \(t_n^3 < 0\) also. Now, let's solve the equation for \(A\). $$ -A^2 = t_n^3 \Rightarrow A^2 = -t_n^3. $$ To get a real solution for \(A\), we require the term inside the square root to be non-negative, that is, \(-t_n^3 \geq 0\). Since \(t_n^3 < 0\), it is always the case that \(-t_n^3 > 0\). Thus, there always exists a value of \(A\) for a negative solution: For every \(t_n < 0\), we have \(t_n^3 < 0\), and thus: $$ A = \pm\sqrt{-t_n^3}. $$

Key Concepts

Solutions of Equations

Whenever we talk about the solutions of an equation, we are essentially discussing the values that make the equation true. For a cubic equation like \(t^3 = -A^2\), the possible "solutions" we are interested in would be the real values that \(t\) can take. In the given exercise, we looked for solutions under specific conditions for \(A\).

Understanding what confirmatory steps are needed, helps in figuring out the relationship between \(t\) and \(A\).

• By substituting \(t = 0\), we determine that the equation becomes \(-A^2 = 0\), leading to \(A^2 = 0\), and finally \(A = 0\).
• To find a positive solution, substituting \(t = t_p > 0\) shows a conflict where \(t_p^3\) is positive but \(-A^2\) remains non-positive. Hence, no solution exists.
• For negative solutions, there is always a corresponding value of \(A\) because substituting \(t = t_n < 0\) aligns in mathematical terms, and gives real values for \(A\).

Understanding these substitutions and their outcomes is the key to determining specific solutions in such equations.

Real Solutions

The concept of real solutions refers to the type of numbers that satisfy an equation in the real number system. Real solutions are distinct from complex or imaginary solutions. In the context of the exercise, we are considering real solutions to the cubic equation \(t^3 = -A^2\).

For a solution to be real, the equation requires the term inside the square root \(\sqrt{-t_n^3}\) to be non-negative. This ensures that we find numbers that exist on the number line without involving imaginary parts.

• The exercise shows that for \(A = 0\), the cubic equation has a real solution \(t = 0\).
• For negative solutions \(t = t_n < 0\), \(A^2 = -t_n^3\) also yields real numbers as values because solving it gives \(A = \pm \sqrt{-t_n^3}\).

Real numbers are straightforward solutions that students can visualize and work with, which are essential for grasping the concept of solutions in algebra.

Negative Solutions

Negative solutions are numbers less than zero that satisfy an equation. In the given problem, finding a negative solution refers to identifying negative values of \(t\) that make the cubic equation \(t^3 = -A^2\) true.

When examining the possibility of a negative solution, substituting \(t = t_n < 0\) helps uncover insights about \(A\). The cubic term \(t_n^3\), when negative, can balance the expression \(-A^2\) provided that \(A\) is a real number.

• The reasoning exposes that since \(t_n^3 < 0\), \(-t_n^3 > 0\). Resolving \(A = \pm \sqrt{-t_n^3}\) confirms the existence of appropriate values for \(A\).
• This illustrates how such equations can comfortably have negative solutions while still adhering to the constraints established by real-number algebra.

Negative solutions challenge intuition initially but reveal a structured path to answering the underlying mathematical inquiries.