Problem 46 - Algebra for College Students

Question

Solve each equation. You will need to use the factoring techniques that we discussed throughout this chapter. $$x^{4}-9 x^{2}=0$$

Step-by-Step Solution

Answer

The solutions are \(x = 0\), \(x = 3\), and \(x = -3\).

Step 1: Identify the common factor

The given equation is \(x^4 - 9x^2 = 0\). Notice that \(x^2\) is a common factor in both terms. This allows us to factor it out: \(x^2(x^2 - 9) = 0\).

Step 2: Set each factor equal to zero

After factoring, set each factor equal to zero. We have two factors: \(x^2 = 0\) and \(x^2 - 9 = 0\). So, we will solve each equation separately.

Step 3: Solve each equation

1. For \(x^2 = 0\), take the square root of both sides to get \(x = 0\). 2. For \(x^2 - 9 = 0\), add 9 to both sides to get \(x^2 = 9\). Then take the square root of both sides to find \(x = 3\) and \(x = -3\).

Step 4: State the solution

The solutions from solving the equations are \(x = 0\), \(x = 3\), and \(x = -3\). These are the values of \(x\) that satisfy the original equation.

Key Concepts

Quadratic Equations

Quadratic equations are equations of the form \(ax^2 + bx + c = 0\). They are called quadratic because 'quad' means square, which refers to the highest degree of the variable \(x\) being squared. In the original exercise, we've already got it factored to something similar in the form \((x^2 - 9)\), which is a bit different because it's missing the middle \(bx\) term. This still falls under the umbrella of quadratic because of the squared factor. Solving these equations usually involves methods such as:

• Factoring, as we've seen in the exercise
• Using the quadratic formula: \(x = \frac{{-b \pm \sqrt{{b^2-4ac}}}}{{2a}}\)
• Completing the square

In the exercise, we just had to find the roots (solutions) by setting each factor equal to zero and solving, which is similar to finding roots of standard quadratic equations.

Polynomial Equations

Polynomial equations are expressions that involve variables raised to any power, such as \(x^4 - 9x^2 = 0\) in our exercise. Each term in a polynomial equation is composed of a coefficient and a variable raised to a non-negative integer power, like 4 and 2 in our problem.

• The degree of the polynomial is the highest power of the variable, which in this example is 4.
• Polynomial equations can often be factored into simpler expressions, which is the technique we used in this exercise.

By recognizing common factors, such as \(x^2\) in this problem, we can transform the equation into a product of factors, making it easier to solve by finding solutions for when each factor equals zero. This breaks down complex polynomial equations into manageable quadratic forms or even simpler problems.

Square Roots

Square roots are operations that, when applied to a number, give a value which multiplied by itself returns the original number. In terms of equations, when we have \(x^2 = a\), taking the square root of both sides helps find \(x\). For example:

• In \(x^2 = 9\), taking the square root gives solutions \(x = 3\) and \(x = -3\).
• Similarly, \(x^2 = 0\) gives \(x = 0\).

When dealing with quadratic or polynomial equations, the square root function can help solve for variable values that satisfy the equation. Always remember that square roots can yield both positive and negative solutions except in cases where they're zero. This is why we had two solutions for our equation besides zero. The square root method is particularly useful in simplifying quadratic forms derived from polynomial equations.