Problem 79 - College Algebra with Modeling and Visualization

Question

Solve for the specified variable. $$ s=-16 t^{2}+100 t \text { for } t $$

Step-by-Step Solution

Answer

\( t = \frac{100 \pm \sqrt{10000 - 64s}}{32} \)

Step 1: Identify the type of equation

The given equation is \( s = -16t^2 + 100t \). This is a quadratic equation in the form of \( at^2 + bt + c = 0 \), where \( a = -16 \), \( b = 100 \), and \( c = -s \).

Step 2: Use the quadratic formula

The quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to solve for \( t \). Here, substituting the values of \( a \), \( b \), and \( c \), we get: \( t = \frac{-100 \pm \sqrt{100^2 - 4(-16)(-s)}}{2(-16)} \).

Step 3: Simplify under the square root

Calculate the discriminant: \( 100^2 - 4(-16)(-s) = 10000 - 64s \). Substitute back into the formula: \( t = \frac{-100 \pm \sqrt{10000 - 64s}}{-32} \).

Step 4: Solve for both solutions

The equation will yield two potential solutions for \( t \): \( t_1 = \frac{-100 + \sqrt{10000 - 64s}}{-32} \) and \( t_2 = \frac{-100 - \sqrt{10000 - 64s}}{-32} \).

Step 5: Simplify the expressions

By simplifying the expressions, you can find the values of \( t \): \[ t_1 = \frac{100 - \sqrt{10000 - 64s}}{32} \] \[ t_2 = \frac{100 + \sqrt{10000 - 64s}}{32} \] Hence, \( t = \frac{100 \pm \sqrt{10000 - 64s}}{32} \).

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. In our scenario, the equation is in the form of \( at^2 + bt + c = 0 \), where each letter represents a numerical coefficient. The formula itself is given as:

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

To apply this formula, identify the coefficients \( a \), \( b \), and \( c \) from your quadratic equation. In our original problem, these coefficients are \( a = -16 \), \( b = 100 \), and \( c = -s \). Inserting these values into the quadratic formula allows you to solve for \( t \).
Use the formula to find two solutions (due to the \( \pm \)) that represent possible solutions for the variable.

Discriminant

The discriminant is a special part of the quadratic formula that determines the nature of the roots. In the formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), the expression under the square root, \( b^2 - 4ac \), is known as the discriminant.
In our exercise, the discriminant becomes \( 100^2 - 4(-16)(-s) \), simplifying to \( 10000 - 64s \). The value of the discriminant tells us about the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it's zero, there is exactly one real root (a repeated root).
• If it's negative, no real roots exist; the solutions are complex or imaginary.

Understanding the discriminant helps us anticipate the number and type of solutions without fully solving the equation.

Solving for a Variable

In the context of a quadratic equation, solving for a variable involves using the quadratic formula to find the value(s) of the unknown that satisfy the equation. Here, our task was to solve for \( t \).
The process starts with identifying the equation as quadratic and then using the quadratic formula to calculate possible solutions. As seen in the step-by-step solution, substituting values into the formula results in two expressions:

• \( t_1 = \frac{100 - \sqrt{10000 - 64s}}{32} \)
• \( t_2 = \frac{100 + \sqrt{10000 - 64s}}{32} \)

These expressions represent the potential solutions for \( t \). Simplifying the expressions further, by calculating the square root and division, gives us the final answer. This is an important part of solving equations, allowing us to find the precise values for the variable in question.