Problem 12 - Algebra and Trigonometry

Question

Determine whether the given value is a solution of the equation. \(x^{2}-b x+\frac{1}{4} b^{2}=0\) (a) \(x=\frac{b}{2} \quad\) (b) \(x=\frac{1}{b}\)

Step-by-Step Solution

Answer

(a) Yes, (b) No.

Step 1: Understand the Equation

The equation given is \( x^2 - bx + \frac{1}{4}b^2 = 0 \). It represents a quadratic equation in the standard form \( ax^2 + bx + c = 0 \). We are tasked with checking if the given values for \( x \) satisfy this equation.

Step 2: Test Value (a) \( x = \frac{b}{2} \)

Substitute \( x = \frac{b}{2} \) into the equation and simplify. \[\left( \frac{b}{2} \right)^2 - b \left( \frac{b}{2} \right) + \frac{1}{4}b^2 = 0\]Calculate each term:- \( \left( \frac{b}{2} \right)^2 = \frac{b^2}{4} \)- \( b \left( \frac{b}{2} \right) = \frac{b^2}{2} \)- \( \frac{1}{4}b^2 \) Substitute and simplify:\[\frac{b^2}{4} - \frac{b^2}{2} + \frac{1}{4}b^2 = 0\]Combine like terms:\[\frac{b^2}{4} + \frac{1}{4}b^2 - \frac{b^2}{2} = \frac{b^2}{2} - \frac{b^2}{2} = 0\] The equation holds true, so \( x = \frac{b}{2} \) is a solution.

Step 3: Test Value (b) \( x = \frac{1}{b} \)

Substitute \( x = \frac{1}{b} \) into the equation and simplify. \[\left( \frac{1}{b} \right)^2 - b \left( \frac{1}{b} \right) + \frac{1}{4}b^2 = 0\]Calculate each term:- \( \left( \frac{1}{b} \right)^2 = \frac{1}{b^2} \)- \( b \left( \frac{1}{b} \right) = 1 \)- \( \frac{1}{4}b^2 \) Substitute and simplify:\[\frac{1}{b^2} - 1 + \frac{1}{4}b^2 = 0\]Combine like terms and rearrange:\[\frac{1}{b^2} - \frac{1}{4}b^2 = 1\]The left side cannot be simplified to equal 1 when \( b eq 0 \), so \( x = \frac{1}{b} \) is not a solution.

Key Concepts

Completing the Square

Completing the square is a technique used to solve quadratic equations, make the expressions easier to understand, or reveal specific properties of the function. This method transforms a quadratic equation into a perfect square trinomial, which can be more intuitive to solve.

Consider the equation from our example: \[x^2 - bx + \frac{1}{4}b^2 = 0\]This already resembles a perfect square trinomial format. To complete the square when faced with a quadratic \(x^2 + px + q\), follow these steps:

• Divide the linear term coefficient \(p\) by 2 and square it. This yields \((\frac{p}{2})^2\).
• Add and subtract this new value inside the equation, if not already present. In our case, we see \(x^2 - bx + \frac{1}{4}b^2\) equals \((x - \frac{b}{2})^2\).

Recognizing the completed square helps to quickly test values like \(x = \frac{b}{2}\), as the equation reduces to zero when this substitution maintains balance.

Solution of Equations

Solving equations involves finding the value of the variable that satisfies the given mathematical statement. In particular, for quadratic equations like \(x^2 - bx + \frac{1}{4}b^2 = 0\), several methods can be used to deduce solutions.

A popular way is by factoring. Notice that the above equation is equivalent to \((x - \frac{b}{2})^2 = 0\). Once factored, solving for \(x\) becomes straightforward:

• Set the inside of the squared term equal to zero: \(x - \frac{b}{2} = 0\).
• Solve for \(x\), yielding \(x = \frac{b}{2}\).

This solution emerges because squaring zero is the only way to achieve a zero product when dealing with squared terms. As the example demonstrates, testing other possible values like \(x = \frac{1}{b}\) verifies they do not satisfy all terms to produce a balanced equation.

Algebraic Substitution

Algebraic substitution is a powerful technique where specific values replace variables in an expression to simplify or solve the equation. In our given problem, we substitute suggested values of \(x\) into the quadratic equation to check for equality.

Here's how substitution aids in testing solutions:

• Choose a provisional value for \(x\), like \(x = \frac{b}{2}\). Substitute it into each occurrence of \(x\) in the equation \(x^2 - bx + \frac{1}{4}b^2\).
• Simplify the expression to determine if both sides of the equation are equal, confirming if it's a solution.
• Repeat with any alternative values provided, such as \(x = \frac{1}{b}\), to see if they too satisfy the equation.

This step-by-step evaluation through substitution not only checks if a particular value solves the equation but also reinforces understanding of how changes in \(x\) affect the equation's balance.