Problem 29 - Algebra A Combined Function

Question

Solve. See Examples 1 through 5. $$ a^{4}-5 a^{2}+6=0 $$

Step-by-Step Solution

Answer

The solutions are \( a = \pm \sqrt{2} \) and \( a = \pm \sqrt{3} \).

Step 1: Substitute Variable

Let us set a new variable for simplification. Let \( x = a^2 \). The original equation \( a^4 - 5a^2 + 6 = 0 \) can now be rewritten as \( x^2 - 5x + 6 = 0 \).

Step 2: Factor the Quadratic Equation

Now, we solve the simplified equation \( x^2 - 5x + 6 = 0 \) by factoring. We look for two numbers that multiply to 6 and add up to -5. These numbers are -2 and -3. This allows us to factorize the equation into \( (x - 2)(x - 3) = 0 \).

Step 3: Solve for x

Using the zero-product property, solve the factored equation \( (x - 2)(x - 3) = 0 \). This gives us two possible solutions: \( x - 2 = 0 \) which yields \( x = 2 \) and \( x - 3 = 0 \) which yields \( x = 3 \).

Step 4: Substitute Back to Find a

Now, substitute \( x = a^2 \) back to find the values of \( a \). For \( x = 2 \), we have \( a^2 = 2 \), which gives roots \( a = \sqrt{2} \) and \( a = -\sqrt{2} \). Similarly, for \( x = 3 \), we have \( a^2 = 3 \), giving roots \( a = \sqrt{3} \) and \( a = -\sqrt{3} \).

Step 5: Solutions Verification

Verify the roots obtained by substituting back to the original equation: \( a = \sqrt{2}, -\sqrt{2}, \sqrt{3}, -\sqrt{3} \) all satisfy \( a^4 - 5a^2 + 6 = 0 \). Calculating \( a^{4}-5 a^{2}+6 \) for each confirms they all simplify to zero.

Key Concepts

Factoring

Factoring is a key technique used to simplify and solve quadratic equations. It involves rewriting a quadratic expression as a product of two binomials. To factor a quadratic equation such as \( x^2 - 5x + 6 = 0 \), we need to find two numbers that multiply to the constant term (in this case, 6) and add up to the linear coefficient (in this case, -5). In this example, -2 and -3 satisfy these conditions: they multiply to 6 and add to -5. Hence, the quadratic can be expressed as \((x - 2)(x - 3) = 0\). This step breaks down complex equations into simpler factors that can be easily solved for variable values.

Zero-product property

The zero-product property is a fundamental concept utilized when solving factored quadratic equations. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. For example, if we have the equation \((x - 2)(x - 3) = 0\), the zero-product property tells us that either \(x - 2 = 0\) or \(x - 3 = 0\) must be true. Therefore, the solutions for \(x\) are 2 and 3. This property is essential in converting a factored expression into straightforward solutions, aiding in solving the entire quadratic equation seamlessly.

Substitution method

The substitution method plays a pivotal role in simplifying and solving polynomial equations, especially when dealing with variables raised to powers higher than two. By assigning \( x = a^2 \), a higher degree equation like \( a^4 - 5a^2 + 6 = 0 \) can be simplified to a quadratic form \( x^2 - 5x + 6 = 0 \). This transformation simplifies the problem, making it easier to solve using factoring. After finding the solutions for \( x \), reverting back to the original variable involves substituting \( x \) with \( a^2 \), which helps in determining the real solutions for the variable \( a \). Substitution is crucial when the original equation is not easily factorable without manipulation.

Polynomial equations

Polynomial equations are mathematical expressions involving variables raised to whole number powers and coefficients. In general, they take the form \( a_nx^n + a_{n-1}x^{n-1} + \, ... \, + a_1x + a_0 = 0 \), where \( n \) is the highest power of the variable in the equation. The equation \( a^4 - 5a^2 + 6 = 0 \) is an example of a polynomial equation. Such equations can often be complex to solve, but techniques like factoring, substitution, and utilizing properties like the zero-product property simplify the process. Solving polynomial equations is all about breaking down the expressions into simpler, manageable parts, thereby making it easier to find all possible solutions.