Problem 36 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ 2 x^{4}-10 x^{2}+8=0 $$

Step-by-Step Solution

Answer

The solutions for the given equation are \( x = 2, -2, 1, -1 \).

Step 1: Substitute a Variable

Let us perform a substitution to simplify the equation. Set \( y = x^2 \). The equation becomes \( 2y^2 - 10y + 8 = 0 \). Our goal is now to solve this quadratic equation in terms of \( y \).

Step 2: Solve the Quadratic Equation

Next, apply the quadratic formula to solve for \( y \). The quadratic formula is \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = -10 \), and \( c = 8 \). Substitute these values into the formula to get:\[ y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4 \times 2 \times 8}}{2 \times 2} \]Calculate inside the square root:\[ = \frac{10 \pm \sqrt{100 - 64}}{4} \]\[ = \frac{10 \pm \sqrt{36}}{4} \]\[ = \frac{10 \pm 6}{4} \]

Step 3: Compute Values for y

Solve for \( y \) by computing the two possible values:1. \( y = \frac{10 + 6}{4} = \frac{16}{4} = 4 \)2. \( y = \frac{10 - 6}{4} = \frac{4}{4} = 1 \)So we have two potential solutions for \( y \), which are \( y = 4 \) and \( y = 1 \).

Step 4: Re-substitute to Solve for x

Since \( y = x^2 \), we solve for \( x \) by equating each value of \( y \) to \( x^2 \):1. \( x^2 = 4 \) gives: - \( x = \sqrt{4} = 2 \) - \( x = -\sqrt{4} = -2 \)2. \( x^2 = 1 \) gives: - \( x = \sqrt{1} = 1 \) - \( x = -\sqrt{1} = -1 \)Thus, the solutions for \( x \) are \( x = 2, -2, 1, -1 \).

Key Concepts

Substitution Method

The substitution method is a powerful tool in algebra that simplifies solving complex equations. By introducing a new variable, we transform the original equation into a more manageable form. For instance, with the equation \(2x^4 - 10x^2 + 8 = 0\), we substitute \(y = x^2\). This makes the equation \(2y^2 - 10y + 8 = 0\).

• This change reduces the complexity and transforms a higher-degree polynomial into a quadratic equation.
• Substitution is often useful when dealing with equations involving even powers of the variable, as it directly reduces the number of degrees.

Using substitution helps us focus on solving a simpler equation first before dealing with the original variable.

Quadratic Formula

The quadratic formula is a universal tool in algebra for solving any quadratic equation of the form \(ax^2 + bx + c = 0\). The formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) finds the roots of the equation by considering the coefficients \(a\), \(b\), and \(c\). In our example, once substituting for \(y\), the quadratic equation \(2y^2 - 10y + 8 = 0\) is solved using:

• Substituting the values \(a = 2\), \(b = -10\), and \(c = 8\).
• Calculating the discriminant \( b^2 - 4ac \) to determine the nature of the roots.
• Evaluating the square root and solving for \(y\).

The quadratic formula helps pinpoint exact solutions efficiently when simple factoring isn't possible.

Algebraic Equations

Algebraic equations denote relationships between variables and constants expressed through mathematical expressions. Solving these equations involves finding the value(s) of the variable(s) that hold the equation true.

• Equations can be linear, quadratic, polynomial, or of higher degrees based on the highest power of the variable.
• Each type of equation might require a different method or formula for solving it.

Understanding the structure of an algebraic equation guides us toward the best strategy for finding solutions, whether it be substitution, using formulas, or other algebraic techniques.

Solutions to Quadratic Equations

Finding solutions to quadratic equations involves determining the values of the variable where the equation equals zero. Using our substitution and quadratic formula, the quadratic equation \(2y^2 - 10y + 8 = 0\) produces solutions \(y = 4\) and \(y = 1\). We translate these back to \(x\) by recognizing \(y = x^2\). This leads to:

• For \(y = 4\), \(x^2 = 4\) implies \(x = 2\) or \(x = -2\).
• For \(y = 1\), \(x^2 = 1\) implies \(x = 1\) or \(x = -1\).

In the context of the original equation, the solutions \(x = 2, -2, 1, -1\) are found, illustrating the connection between the solved quadratic and our initial higher degree polynomial. This process showcases the elegance of algebra in resolving complex equations to simple solutions.