Problem 71 - Beginning and Intermediate Algebra

Question

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$45 h=20 h^{3} \quad\( 72) \)64 d^{3}=100 d$$

Step-by-Step Solution

Answer

The short answer: For the first equation, the solutions are \(h = 0\), \(h = -\frac{3}{2}\), and \(h = \frac{3}{2}\). For the second equation, the solutions are \(d = 0\), \(d = -\frac{5}{4}\), and \(d = \frac{5}{4}\).

Step 1: Equation 1: Factoring and Applying the Zero Product Rule

Given the equation \(45h = 20h^3\), first move all terms to one side of the equation: $$20h^3 - 45h = 0$$ Now, we can factor out the common factor of h: $$h(20h^2 - 45) = 0$$

Step 2: Equation 1: Solving for h

Now by the zero product rule, one of the factors must be equal to zero: Case 1: If \(h = 0\), the equation is satisfied. Case 2: If the second factor is zero: $$20h^2 - 45 = 0$$ Now solve for h: $$20h^2 = 45$$ $$h^2 = \frac{45}{20}$$ $$h = \pm\sqrt{\frac{45}{20}}$$ $$h = \pm\sqrt{\frac{9}{4}}$$ $$h = \pm\frac{3}{2}$$ So the solutions for the first equation are: \(h = 0\), \(h = -\frac{3}{2}\), and \(h = \frac{3}{2}\).

Step 3: Equation 2: Factoring and Applying the Zero Product Rule

Given the equation \(64d^3 = 100d\), first move all terms to one side of the equation: $$64d^3 - 100d = 0$$ Now, we can factor out the common factor of d: $$d(64d^2 - 100) = 0$$

Step 4: Equation 2: Solving for d

Now by the zero product rule, one of the factors must be equal to zero: Case 1: If \(d = 0\), the equation is satisfied. Case 2: If the second factor is zero: $$64d^2 - 100 = 0$$ Now solve for d: $$64d^2 = 100$$ $$d^2 = \frac{100}{64}$$ $$d = \pm\sqrt{\frac{100}{64}}$$ $$d = \pm\sqrt{\frac{25}{16}}$$ $$d = \pm\frac{5}{4}$$ So the solutions for the second equation are: \(d = 0\), \(d = -\frac{5}{4}\), and \(d = \frac{5}{4}\).

Key Concepts

Factoring

When you come across an equation like the ones in the example, sometimes the main goal is to simplify it by expressing it as a product of simpler terms. This process is known as factoring.

Imagine you start with an equation such as \(20h^3 - 45h = 0\). The first step involves identifying common factors among the terms. In this example, both terms have the variable \(h\) in common.

By factoring out \(h\), the expression becomes \(h(20h^2 - 45) = 0\). Now the equation is simplified into a product of factors. That makes it easier to apply further methods to solve it.

• Always look for common factors first.
• Factoring reduces a complex equation into simpler expressions.
• The expression should always equal zero before factoring to use the Zero Product Rule effectively.

Zero Product Rule

After you successfully factor the equation, the Zero Product Rule becomes useful. This rule states that if a product of several factors equals zero, at least one of the factors themselves must be zero.

This principle is what allows you to set each factor in an equation to zero separately. For example, from \(h(20h^2 - 45) = 0\), it follows that either \(h = 0\) or \(20h^2 - 45 = 0\).

• Only use the Zero Product Rule when the equation is set to zero.
• You solve for variables by evaluating each factor separately.
• This method can reveal multiple solutions for the equation.

Applying the Zero Product Rule is straightforward but very powerful, especially for polynomial equations like the ones here.

Solving Equations

Once you have applied the Zero Product Rule, you are left with simpler equations to solve. Let’s say, in the factoring step, you ended up with the equations \(h = 0\) and \(20h^2 - 45 = 0\). Solving \(h = 0\) is immediate, as \(h\) is already defined.

For the second equation, you'll need to isolate the variable. Rearrange \(20h^2 = 45\) by dividing both sides by 20, resulting in \(h^2 = \frac{45}{20}\). Further simplify the fraction to give \(h^2 = \frac{9}{4}\).

Finding \(h\) results in taking the square root. Here, \(h = \pm\frac{3}{2}\). This shows that one step leads to multiple solutions, highlighting the possibilities when using algebraic methods.

• Simplify each equation to isolate variables.
• Consider square roots or other operations carefully.
• The solutions represent points at which the original equations are satisfied.

Remember that solving equations involves multiple steps that align in a sequence: simplify, isolate, and solve. Mastery of these skills can be highly beneficial.