Problem 39 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ x^{4}-13 x^{2}+40=0 $$

Step-by-Step Solution

Answer

The real solutions are \( x = \pm 2\sqrt{2} \) and \( x = \pm \sqrt{5} \).

Step 1: Observe the Structure

The given equation is a polynomial equation in the form of a quartic (degree 4) polynomial: \[ x^4 - 13x^2 + 40 = 0 \]The equation is structured as a quadratic in terms of \( x^2 \). Let's use substitution to simplify it.

Step 2: Substitute to Simplify

Let us substitute \( y = x^2 \). The equation now becomes: \[ y^2 - 13y + 40 = 0 \]Now, we need to solve this quadratic equation for \( y \).

Step 3: Solve the Quadratic Equation

To solve \( y^2 - 13y + 40 = 0 \), we will use the quadratic formula: \[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -13 \), and \( c = 40 \). Calculate the discriminant:\[ b^2 - 4ac = (-13)^2 - 4 \times 1 \times 40 = 169 - 160 = 9 \]Thus, the solutions for \( y \) become:\[ y = \frac{13 \pm 3}{2} \]This yields two values: \( y = 8 \) and \( y = 5 \).

Step 4: Reverse Substitution

Since we substituted \( y = x^2 \), we now revert back to \( x \). We have two equations now:1. \( x^2 = 8 \)2. \( x^2 = 5 \)Take the square root of each:1. \( x^2 = 8 \) gives \( x = \pm \sqrt{8} = \pm 2\sqrt{2} \)2. \( x^2 = 5 \) gives \( x = \pm \sqrt{5} \)

Step 5: Write the Real Solutions

The real solutions for \( x \) are derived from the square roots as calculated:\( x = 2\sqrt{2}, -2\sqrt{2}, \sqrt{5}, -\sqrt{5} \)Therefore, the complete solution set is: \[ x = \pm 2\sqrt{2} \quad \text{and} \quad x = \pm \sqrt{5} \]

Key Concepts

Real Solutions

When it comes to solving equations, particularly polynomial ones, we often seek what are called "real solutions." These are the values of the variable, in this case, \( x \), that satisfy the equation and are not imaginary or complex.
Real solutions are typically found within the set of real numbers, which include all positive numbers, negative numbers, and zero.
In the example of the quartic equation \( x^4 - 13x^2 + 40 = 0 \), real solutions would be the values of \( x \) that make this equation true.In general, finding real solutions involves several steps:

• Substituting to simplify the equation, if necessary.
• Solving for the variable after substituting.
• Interpreting the results to identify the real solutions.

Real solutions are immensely important as they often represent actual quantities or measurements in real-world scenarios.

Quadratic Substitution

The method of quadratic substitution is a powerful tool to simplify complex polynomial equations into a form that is more manageable.
In particular, for the quartic equation \( x^4 - 13x^2 + 40 = 0 \), we use substitution by recognizing the term \( x^2 \) as a single entity.
Here's how it works:

• Identify that the equation is structured as a quadratic in terms of \( x^2 \).
• Introduce a new variable, let's say \( y = x^2 \), which transforms the quartic equation into a simpler quadratic equation \( y^2 - 13y + 40 = 0 \).
• By solving this new quadratic equation, we can create a pathway back to the original variable \( x \).

Quadratic substitution is particularly helpful because it can reduce an otherwise cumbersome deeper degree polynomial problem into one that can be tackled using more straightforward methods like the quadratic formula.

Polynomial Equations

Polynomial equations are equations made up of terms that involve only non-negative integer powers of the variable.
These equations form the backbone of algebra and have a wide range of complexity based on the degree of the polynomial.Here are some key points about polynomial equations:

• The degree of the polynomial is determined by the highest power of the variable, as seen in \( x^4 \) indicating a quartic polynomial.
• The structure of polynomial equations can often lead to recognizing patterns or forms suitable for substitution, like in our example where \( x^4 \) was expressed in terms of \( x^2 \).
• Solutions to polynomial equations can be real or complex, and the number of solutions is typically equal to the degree of the equation.

Understanding polynomial equations is crucial as they appear in various fields including physics, engineering, and economics, representing phenomena that follow a specific pattern or rule.

Discriminant Calculation

A discriminant in a quadratic equation provides vital information about the nature of its solutions.
The quadratic formula for checking solutions is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The part under the square root, \( b^2 - 4ac \), is the discriminant.For the equation \( y^2 - 13y + 40 = 0 \), let's see how the discriminant aids in predicting the type of solutions:

• If the discriminant is positive, the equation has two distinct real solutions, as seen here where \( b^2 - 4ac = 9 \), a positive number.
• If the discriminant is zero, there is one real solution, indicating a perfect square trinomial.
• If the discriminant is negative, the solutions are complex or imaginary, meaning they lie outside the real number system.

Thus, calculating the discriminant helps us not only solve the equation efficiently but also understand the characteristics of the solutions we are dealing with.