Problem 48 - Algebra 1: Concepts and Skills

Question

Solve the equation. Tell which method you used. \(5 x^{4}-80 x^{2}=0\)

Step-by-Step Solution

Answer

The roots of the given equation are 0, -4, and 4.

Step 1: Rewriting the Equation

Firstly, rewrite the given equation \(5 x^{4}-80 x^{2}=0\) in terms of \(x^2\), which we can denote as \(y\). This gives us \(5y^2-80y=0\).

Step 2: Factor Out the Common Factor

In the rewritten equation, the common factor from both terms is the denominator, which is 5y. Factoring this out gives \(5y(y-16)=0\).

Step 3: Apply the Zero-Product Property

According to the zero-product property, if a product of factors equals zero, then at least one of the factors must be zero. So, set y to 0 and y - 16 to 0 respectively. This gives two values for y: 0 and 16.

Step 4: Substitute y with \(x^2\)

Recall that y was a placeholder for \(x^2\). So replace y in the roots with \(x^2\). This gives \(x^2 = 0\) or \(x^2 = 16\).

Step 5: Solve for x

To find the values for x, take the square root of both equations. From \(x^2 = 0\), \(x\) can be 0. From \(x^2 = 16\), \(x\) can be 4 or -4.

Key Concepts

Zero-Product Property

The zero-product property is a fundamental principle used in algebra for solving equations. It states that if the product of two numbers is zero, at least one of the numbers must be zero. This is highly useful when dealing with polynomial equations.
For example, consider the equation that has been factored as \(a \cdot b = 0\). According to the zero-product property:

• Either \(a = 0\)
• Or \(b = 0\)

In our original equation \(5y(y-16)=0\), applying this property means setting each factor equal to zero:

• \(5y = 0\)
• \(y - 16 = 0\)

These lead to two simple equations that can be solved individually.

Factoring Equations

Factoring is a method used to break down complex algebraic expressions into simpler components called factors. It is a crucial step when solving polynomial equations, especially when used in conjunction with the zero-product property.
In our case of the equation \(5 x^{4}-80 x^{2}=0\), the first step is to identify a common factor in all terms. Here, \(5x^2\) is the common factor. By factoring out \(5x^2\), the equation becomes easier to manage: \[5x^2 (x^2 - 16)=0\]
Factoring further, recognize \(x^2 - 16\) as a difference of squares:\[x^2 - 16 = (x - 4)(x + 4)\] Now, the expression is fully factored into \(5x^2(x-4)(x+4)=0\), a crucial step before applying the zero-product property.

Solving Polynomial Equations

Solving polynomial equations involves a series of strategic steps aimed at finding the variable values that satisfy the equation. These values are called the 'roots' of the equation.
Let's revisit our exercise. First, we needed to simplify and factor the equation:

• Initial equation: \(5 x^{4}-80 x^{2}=0\)
• After substitution and factoring: \(5x^2(x-4)(x+4)=0\)

Once shown as a product of factors, apply the zero-product property to set each factor to zero.

• Set \(5x^2 = 0\): Solve to find \(x = 0\).
• Set \(x-4 = 0\): Solution is \(x = 4\).
• Set \(x+4 = 0\): Solution is \(x = -4\).

The complete set of solutions for the equation is \(x \in \{ 0, 4, -4 \}\). Achieving this solution involves understanding both the factoring and application of the zero-product property, showcasing the structured problem-solving that polynomial equations require.