Problem 34 - Intermediate Algebra

Question

Solve each equation. $$ x^{4}-21 x^{2}+54=0 $$

Step-by-Step Solution

Answer

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

Step 1: Substitute Variables

Let us make the equation easier by substituting a variable. Let \( y = x^2 \). This changes the equation from \( x^4 - 21x^2 + 54 = 0 \) to \( y^2 - 21y + 54 = 0 \).

Step 2: Identify Form of the Equation

The substituted equation \( y^2 - 21y + 54 = 0 \) is a quadratic equation. We need to solve for \( y \) using the quadratic formula.

Step 3: Use the Quadratic Formula

Apply the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1 \), \( b = -21 \), and \( c = 54 \). Substitute these values to get:\[y = \frac{-(-21) \pm \sqrt{(-21)^2 - 4\times1\times54}}{2\times1} = \frac{21 \pm \sqrt{441 - 216}}{2} = \frac{21 \pm \sqrt{225}}{2}.\]

Step 4: Calculate Solutions for y

Simplify the square root to find \( y \):\[y = \frac{21 \pm 15}{2}.\]This gives us two potential solutions for \( y \):\( y_1 = \frac{36}{2} = 18 \) and \( y_2 = \frac{6}{2} = 3 \).

Step 5: Back-Substitute to Solve for x

Recall that \( y = x^2 \), so solve for \( x \) by substituting each value of \( y \):For \( y = 18 \):\[x^2 = 18 \Rightarrow x = \pm \sqrt{18} = \pm 3\sqrt{2}.\]For \( y = 3 \):\[x^2 = 3 \Rightarrow x = \pm \sqrt{3}.\]

Step 6: State the Final Solution

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

Key Concepts

Quadratic Equations

Quadratic equations are polynomial equations of the form:

• \[ ax^2 + bx + c = 0 \]

In this form, \( a \), \( b \), and \( c \) are constants, where \( a eq 0 \). Quadratic equations can be solved using various methods such as factoring, completing the square, or using the quadratic formula.
Quadratic equations play a fundamental role in intermediate algebra due to their wide applications and relatively simple structure once recognized. In the given exercise, we transformed a fourth-degree polynomial into a quadratic equation by substitution.
The equation \( y^2 - 21y + 54=0 \) is a simplified quadratic form from the original equation.
With quadratic equations, you can expect typically two solutions (though complex numbers may appear sometimes). Recognizing when and how to use these forms is essential.

Substitution Method

The substitution method is a technique used to simplify complex equations by replacing a variable or expression with a simpler one.
In our exercise, we used substitution to turn a fourth-degree polynomial into a more manageable quadratic equation.

• We started by letting \( y = x^2 \).
• This transformed the equation \( x^4 - 21x^2 + 54 = 0 \) into a quadratic form: \( y^2 - 21y + 54 = 0 \).

The substitution method helps reduce the complexity of the problem, enabling us to utilize standard techniques for solving quadratic equations. Once we solved for \( y \), we back-substituted to find the solutions for \( x \).
This process highlights the versatility and importance of substitution in algebra, making it a vital problem-solving strategy.

Square Roots

Square roots are fundamental in solving equations, especially when dealing with quadratic equations or their solutions.
The square root of a number \( x \) is a value that, when multiplied by itself, gives \( x \). This principle helps undo a power of two, as seen in our solution.

• For \( y = 18 \), we found \( x^2 = 18 \). Therefore, \( x = \pm\sqrt{18} = \pm3\sqrt{2} \).
• For \( y = 3 \), similarly, \( x^2 = 3 \) led to \( x = \pm\sqrt{3} \).

Understanding square roots allows us to determine both positive and negative root solutions, which are crucial in polynomial equations. Solving square roots manually also reinforces comprehension in its calculation and simplification process, as seen here with \( 3\sqrt{2} \). These solutions are essential for grasping further mathematical concepts in algebra and beyond.