Problem 58 - Algebra A Combined Function

Question

Solve each equation. $$ x^{3}-14 x^{2}+49 x=0 $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = 7 \).

Step 1: Factor Out the Common Term

The given equation is \( x^{3} - 14x^{2} + 49x = 0 \). We first notice that each term in the equation has an \( x \) in common, so we can factor \( x \) out: \[x(x^{2} - 14x + 49) = 0\]This step simplifies the equation by reducing the terms inside the parentheses.

Step 2: Identify Solutions from Factored Terms

Now that we have a product of terms where the equation equals zero, we can apply the zero-product property. The zero-product property states that if a product equals zero, then at least one of the factors must be zero. Based on this, we solve:1. \( x = 0 \) (from the factor \( x \)).2. \( x^{2} - 14x + 49 = 0 \) (from the quadratic factor).

Step 3: Solve the Quadratic Equation

The quadratic equation to solve is \( x^{2} - 14x + 49 = 0 \). This is a perfect square trinomial that can be rewritten as:\[(x - 7)^{2} = 0\]This simplifies to \( x - 7 = 0 \), giving us \( x = 7 \).

Step 4: List All Unique Solutions

From the steps above, we obtain the solutions \( x = 0 \) and \( x = 7 \). The solution \( x = 7 \) appears twice due to the perfect square quadratic. Therefore, the distinct solutions are \( x = 0 \) and \( x = 7 \).

Key Concepts

Factoring

Factoring is a process used to simplify expressions or equations by breaking them down into smaller multiplicative components. Think about factoring as a way of finding things that multiply together to create the original expression. In the context of solving equations, it is particularly useful because it can make complex equations easier to solve.

Let's consider the expression from the exercise: \[x^{3} - 14x^{2} + 49x = 0\]

In this example, each term contains an "x." This is a clue that "x" is a common factor. By factoring out the common term "x," we rewrite the equation as:\[x(x^{2} - 14x + 49) = 0\]

• This helps us see each part of the equation more clearly.
• It sets us up for applying other algebraic techniques next.

Factoring not only simplifies equations but also turns them into more accessible forms useful for solving.

Zero-Product Property

The zero-product property is a fundamental principle in algebra. It states that if you multiply two or more factors and the result is zero, then at least one of the factors must be zero. This is really powerful for solving equations because it provides a straightforward method to find solutions.

In the exercise, after factoring out the common "x," we have:

• \(x(x^{2} - 14x + 49) = 0\)

Using the zero-product property, we can conclude:

• Either \(x = 0\) or \(x^{2} - 14x + 49 = 0\).

Each factor that equals zero means a possible solution to our original equation. This makes the zero-product property an essential tool for solving polynomial equations, because it splits a potentially complex problem into simpler parts you can handle individually.

Quadratic Equations

A quadratic equation is any equation that can be written in the standard form \(ax^{2} + bx + c = 0\). Quadratics are special because they often appear in many areas of math, and they are known to have at most two solutions.

From our equation:\[x^{2} - 14x + 49 = 0\]

• This is a quadratic equation with \( a = 1 \), \( b = -14 \), and \( c = 49 \).

Solving a quadratic can be approached in various ways: factoring, using the quadratic formula, or completing the square. Recognizing perfect square trinomials (like in our case) is one of the most straightforward methods. It simplifies understanding the nature of roots or solutions a quadratic equation has.

Perfect Square Trinomials

Perfect square trinomials are special quadratic expressions that can be expressed as the square of a binomial. Recognizing these expressions can make solving quadratic equations much easier.

In the exercise, the quadratic \(x^{2} - 14x + 49 = 0\) is a perfect square trinomial.

• You can rewrite it as \((x - 7)^{2} = 0\).
• This form quickly identifies \(x - 7 = 0\), leading to the solution \(x = 7\).

Understanding perfect squares helps easily spot when a quadratic takes on that form. This simplifies solving by allowing direct deduction of the solution. Such trinomials reinforce the idea that solving can often be achieved through observation and pattern recognition.