Problem 59 - Algebra and Trigonometry

Question

55–64 ? Find all solutions, real and complex, of the equation. $$ x^{4}-6 x^{2}+8=0 $$

Step-by-Step Solution

Answer

Solutions: \( x = \pm \sqrt{2}, \pm 2 \).

Step 1: Substitution

To simplify the equation, we substitute: let \( y = x^2 \). This transforms the equation to \( y^2 - 6y + 8 = 0 \).

Step 2: Factoring the Quadratic

Next, factor the quadratic equation \( y^2 - 6y + 8 = 0 \). This can be factored into \( (y - 2)(y - 4) = 0 \).

Step 3: Solving for y

Set each factor to zero and solve for \( y \):1. \( y - 2 = 0 \Rightarrow y = 2 \)2. \( y - 4 = 0 \Rightarrow y = 4 \)

Step 4: Substitute Back to x

Now, substitute back to \( x^2 = y \):1. \( x^2 = 2 \Rightarrow x = \pm \sqrt{2} \)2. \( x^2 = 4 \Rightarrow x = \pm 2 \)

Step 5: Conclude Solutions

Thus, the solutions to the original equation \( x^4 - 6x^2 + 8 = 0 \) are \( x = \pm \sqrt{2}, \pm 2 \). These are all real numbers.

Key Concepts

Quadratic Substitution

Quadratic substitution is a handy technique used when dealing with higher-degree polynomial equations. It simplifies the equations by reducing the degree, making them easier to solve. In the problem provided, you have a quartic equation, which is a polynomial of degree 4:
\[ x^4 - 6x^2 + 8 = 0 \]
To make things simpler, we introduce a new variable, let's say \( y \), where \( y = x^2 \). This transforms the original quartic equation into a quadratic equation:
\[ y^2 - 6y + 8 = 0 \]
By making this substitution, you turn a potentially complex problem into a more familiar quadratic equation, paving the way for easier methods of solving.

Factoring Quadratics

Factoring is a fundamental method used to solve quadratic equations. Once you have a quadratic equation, like the substituted one:
\[ y^2 - 6y + 8 = 0 \]
the next step is to express it as a product of binomials. This means finding two numbers that multiply to give the constant term (8 in this case) and add up to the coefficient of the linear term (-6). The factors here will be:

• \((y - 2)\)
• \((y - 4)\)

So, the equation factors to:
\[ (y - 2)(y - 4) = 0 \]
This method of factoring provides solutions for \( y \) by setting each factor equal to zero. It's efficient and works well with many quadratic equations.

Real Solutions

Finding real solutions involves identifying the values of \( x \) that satisfy the original equation and are real numbers. After solving the factored quadratic equation:
- \( y = 2 \) - \( y = 4 \)
These solutions are substituted back into \( x^2 = y \) since \( y = x^2 \). Solving these gives:
- For \( y = 2 \): \( x^2 = 2 \Rightarrow x = \pm \sqrt{2} \)- For \( y = 4 \): \( x^2 = 4 \Rightarrow x = \pm 2 \)
All these results are real numbers because they correspond to valid square roots, which is important as sometimes solutions can be complex.

Complex Solutions

Complex solutions arise when solving equations with no real number roots. In this particular exercise, all solutions we found are real. However, understanding complex solutions aids in dealing with similar equations when complex values are involved. Complex numbers are rooted in the idea that the square root of a negative number involves an imaginary unit \( i \), where \( i = \sqrt{-1} \).
For example, if we had a situation where setting \( y \) in \( y^2 + 1 = 0 \) led to \( y = -1 \), solving \( x^2 = -1 \) results in complex solutions: \( x = \pm i \).
Complex solutions play a crucial role mainly in equations where the standard real number system would not provide a solution.