Problem 39 - Algebra 2 and Trigonometry

Question

\(\ln 38-43,\) solve for \(x\) $$ _{x} P_{6}=30\left(_{x} P_{4}\right) $$

Step-by-Step Solution

Answer

The value of \(x\) is 10.

Step 1: Understand the Permutations Formula

Permutation is the arrangement of a subset of items from a larger pool, and is defined by \(_nP_r = \frac{n!}{(n-r)!}\). The '!' denotes factorial, representing the product of an integer and all the integers below it.

Step 2: Apply the Permutations Formula

We have \(_xP_6 = \frac{x!}{(x-6)!}\) and \(_xP_4 = \frac{x!}{(x-4)!}\). Substituting these into the given equation leads to \(\frac{x!}{(x-6)!} = 30 \times \frac{x!}{(x-4)!}\).

Step 3: Simplify the Equation

First, cancel out \(x!\) from both sides since it appears in both the numerator and denominator. This simplifies to \(\frac{1}{(x-6)!} = 30 \times \frac{1}{(x-4)(x-5)(x-6)!}\).

Step 4: Eliminate the Factorials from the Equation

Now, multiplying both sides by \((x-6)!\), we get \(1 = 30 \times \frac{1}{(x-4)(x-5)}\). This simplifies further to 1 = \(\frac{30}{(x-4)(x-5)}\).

Step 5: Solve the Simplified Equation

Rearrange the equation to solve for \((x-4)(x-5) = 30\). Expand this to form a quadratic equation: \(x^2 - 9x + 20 = 30\).

Step 6: Simplify the Quadratic Equation

Rearrange the quadratic equation to standard form: \(x^2 - 9x - 10 = 0\).

Step 7: Factor the Quadratic Equation

Factor the equation: \((x-10)(x+1) = 0\). This gives the solutions for \(x\): \(x = 10\) and \(x = -1\).

Step 8: Verify the Solutions

Since a permutation involving a negative number of items doesn't make sense in the context, discard \(x = -1\). Therefore, the solution is \(x = 10\).

Key Concepts

Factorial Notation

Factorial notation, denoted by an exclamation mark \(n!\), represents the product of all positive integers less than or equal to n. It's a basic yet powerful concept primarily used in permutations and combinations.
For instance, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). Factorials are fundamental in calculating permutations, allowing us to understand arrangements of different subsets from a larger set.
Factorials also appear in other areas of mathematics, including series expansion and probability, making them a crucial concept to master for further studies.

• \(0!\) is defined as 1 by convention.
• Factorials grow very quickly with larger numbers.

Understanding factorial notation is key to effectively solving problems that involve permutations.

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\), where a, b, and c are constants, and \(x\) is the variable. Solving quadratic equations is a staple of algebra that extends to many applications in mathematics and science.
Quadratic equations display a parabolic graph, opening upwards or downwards depending on the coefficient a.
Common methods for solving quadratic equations include:

• Factoring: Expressing the quadratic as a product of two binomials.
• Completing the square: Rewriting the equation so one side forms a perfect square trinomial.
• Quadratic formula: Using the formula \(x = {-b \pm \sqrt{b^2 - 4ac} \over 2a}\) to find \(x\).

Depending on the discriminant \(b^2 - 4ac\), quadratic equations can have zero, one, or two real roots. Understanding these roots is essential in interpreting physical and geometrical problems.

Solving Algebraic Equations

Solving algebraic equations involves finding the values of variables that make the equation true. These equations form the foundation of algebra, covering a wide range of concepts and their applications.
Steps to solving algebraic equations typically include:

• Combining like terms.
• Isolating the variable on one side of the equation.
• Using inverse operations to simplify and solve.

It's crucial to check solutions by substituting them back into the original equation to ensure they satisfy it. Sometimes, a solution may not make sense in context, as seen in permutations where negative solutions are not feasible.
The goal is always to make the equation simpler and more manageable until the answer is clear, reinforcing the principle that algebra is about understanding relationships between numbers and operations.