Problem 56 - Precalculus Mathematics for Calculus

Question

Find all real solutions of the equation, rounded to two decimals. $$x^{4}-8 x^{2}+2=0$$

Step-by-Step Solution

Answer

The solutions are \( x = \pm 2.78 \) and \( x = \pm 0.51 \).

Step 1: Substitute Variables for Simplicity

Let's substitute \( y = x^2 \) to transform the equation. This substitution simplifies \(x^4\) into \( (x^2)^2 = y^2 \). Substitute this into the equation, transforming it to \( y^2 - 8y + 2 = 0 \). This is now a quadratic equation.

Step 2: Solve the Quadratic Equation

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -8 \), and \( c = 2 \) to solve the quadratic equation \( y^2 - 8y + 2 = 0 \). The discriminant \( b^2 - 4ac = 64 - 8 = 56 \), which is positive, ensuring real solutions exist. Calculate:\[ y_1 = \frac{8 + \sqrt{56}}{2} \approx 7.74 \] and \[ y_2 = \frac{8 - \sqrt{56}}{2} \approx 0.26 \].

Step 3: Re-substitute to Find x

Re-substitute \( y = x^2 \) back into the solutions found: \( x^2 = y_1 \) and \( x^2 = y_2 \). Thus, \( x = \pm \sqrt{7.74} \) and \( x = \pm \sqrt{0.26} \).

Step 4: Calculate the x Values

Compute the square roots to find the actual values of \( x \):- For \( x = \pm \sqrt{7.74} \), calculate to get approximately \( x \approx \pm 2.78 \).- For \( x = \pm \sqrt{0.26} \), calculate to get \( x \approx \pm 0.51 \).

Step 5: Consolidate the Solutions

Combine all real solutions from earlier steps. The solutions you obtained will be \( x = 2.78, -2.78, 0.51, -0.51 \).

Key Concepts

Quadratic Equation

A quadratic equation is a fundamental concept in algebra. It is written in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients and \( x \) represents the variable. Quadratic equations are prevalent because they model many real-world scenarios. For instance, projectile motion and calculating areas.There are several ways to solve quadratic equations:

• Factoring
• Completing the square
• Using the quadratic formula

In our exercise, the substitution method is used first to transform a quartic equation into a simpler quadratic equation. By letting \( y = x^2 \), we were able to rewrite the quartic equation as a quadratic equation: \( y^2 - 8y + 2 = 0 \). This makes it more accessible to apply the quadratic formula.

Discriminant

The discriminant is a key part of the quadratic equation's solution. It is found in the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) and is the part under the square root: \( b^2 - 4ac \). This value is crucial because it determines the nature of the roots of the equation.Here’s what different values of the discriminant indicate:

• If the discriminant is positive, there are two distinct real solutions.
• If it is zero, there is exactly one real solution (a repeated root).
• If it is negative, the solutions are not real but complex numbers.

In our problem, the discriminant is calculated as \( 56 \), which is positive. This means the quadratic equation \( y^2 - 8y + 2 = 0 \) has two distinct real solutions.

Substitution Method

The substitution method is a powerful algebraic tool used to simplify complex equations. In the original problem, which involved a quartic equation \( x^4 - 8x^2 + 2 = 0 \), substituting \( y = x^2 \) transformed it into something more familiar: \( y^2 - 8y + 2 = 0 \).This process is beneficial because:

• It simplifies higher-degree polynomial equations.
• Makes the equation easier to solve using known techniques.
• Helps in breaking down problems into manageable parts.

After solving the quadratic equation for \( y \), re-substituting \( y = x^2 \) allows us to find the values of \( x \). This two-step approach neatly handles the complexity of the initial equation.

Real Solutions

Real solutions refer to the solutions of an equation that are numbers on the real number line. These are the solutions where you can calculate an actual number without involving the imaginary unit \( i \), which signifies the square root of \( -1 \).When solving quadratic equations, real solutions come from either a positive or zero discriminant. In our exercise, the quadratic equation \( y^2 - 8y + 2 = 0 \) had a discriminant of \( 56 \), leading to real solutions \( y_1 \approx 7.74 \) and \( y_2 \approx 0.26 \).After determining \( y \), the substitution \( x = \pm \sqrt{y} \) finds the real solutions for \( x \) in the original equation. Thus, from \( y_1 \) and \( y_2 \), we deduced \( x = 2.78, -2.78, 0.51, -0.51 \) as the real solutions of the initial quartic equation.