Problem 74 - Algebra and Trigonometry

Question

Solve the equation for \(x\) $$ b^{2} x^{2}-5 b x+4=0 \quad(b \neq 0) $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{4}{b} \) and \( x = \frac{1}{b} \).

Step 1: Identify the quadratic equation

The given equation is \( b^2 x^2 - 5bx + 4 = 0 \). This is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where \( a = b^2 \), \( b = -5b \), and \( c = 4 \).

Step 2: Use the Quadratic Formula

The quadratic formula is given as \( x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = b^2 \), \( B = -5b \), and \( C = 4 \). Substitute these values into the formula.

Step 3: Substitute into the formula

Substitute the values into the quadratic formula: \[ x = \frac{-(-5b) \pm \sqrt{(-5b)^2 - 4(b^2)(4)}}{2(b^2)} \] This simplifies to: \[ x = \frac{5b \pm \sqrt{25b^2 - 16b^2}}{2b^2} \]

Step 4: Simplify under the square root

Calculate the expression under the square root: \[ 25b^2 - 16b^2 = 9b^2 \] So, the equation becomes: \[ x = \frac{5b \pm \sqrt{9b^2}}{2b^2} \]

Step 5: Evaluate the square root

Calculate the square root: \( \sqrt{9b^2} = 3b \). Substitute this back into the equation to get: \[ x = \frac{5b \pm 3b}{2b^2} \]

Step 6: Solve for both solutions

Consider both the plus and minus cases in the quadratic formula: 1. \( x = \frac{5b + 3b}{2b^2} = \frac{8b}{2b^2} = \frac{4}{b} \)2. \( x = \frac{5b - 3b}{2b^2} = \frac{2b}{2b^2} = \frac{1}{b} \)

Step 7: Conclusion

The solutions to the quadratic equation \( b^2 x^2 - 5bx + 4 = 0 \) are \( x = \frac{4}{b} \) and \( x = \frac{1}{b} \).

Key Concepts

Quadratic Formula

The quadratic formula is a fundamental tool in algebra for solving quadratic equations. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). The coefficients \( a \), \( b \), and \( c \) represent numerical values. The quadratic formula is expressed as:\[x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\]This formula allows you to find the solutions, or roots, of the quadratic equation. These solutions could be real or complex numbers, depending on the value inside the square root, known as the discriminant (\( B^2 - 4AC \)).
Certain discriminant values lead to different types of solutions:

• If the discriminant is greater than zero, there are two distinct real solutions.
• If it equals zero, there's exactly one real solution, often referred to as a repeated root.
• If it's less than zero, the solutions are complex or imaginary numbers.

Understanding the quadratic formula is key to tackling quadratic problems efficiently.

Solving Equations

Solving equations involves finding the value of unknown variables that satisfy the equation. In the context of quadratic equations like \( b^2 x^2 - 5bx + 4 = 0 \), we can solve for \( x \) by using the quadratic formula.
The main steps in solving quadratic equations include:

• Identifying the coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
• Plugging these coefficients into the quadratic formula.
• Simplifying the expression, especially under the square root, to make calculating easier.
• Evaluating both the positive and negative results of the "plus-minus" symbol \( \pm \) in the formula, which stands for two possible solutions.

This process helps in deriving the possible values for \( x \) that make the original equation true.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In equations such as \( b^2 x^2 - 5bx + 4 = 0 \), algebraic expressions include terms like \( b^2x^2 \), which signifies "b squared times x squared."
Algebraic expressions serve as foundational building blocks in algebra. They are useful for representing real-world problems and relationships in mathematical forms.
When working with algebraic expressions in quadratic equations:

• Identify each term clearly, noting the coefficients and powers of the variables.
• Simplify expressions by combining like terms or factoring when possible.
• Understand how each term contributes to the overall solution of the equation.

By understanding and manipulating these expressions, we solve for unknowns in different types of equations effectively, revealing the possibilities proposed by the problem.