Problem 56 - Algebra and Trigonometry with Analytic Geometry

Question

Determine values for \(a\) and \(b\) such that \(\frac{5}{3}\) is a solution of the equation. $$a x^{2}+b x=0$$

Step-by-Step Solution

Answer

One possible set of coefficients is \( a = 3 \) and \( b = -5 \).

Step 1: Substitute Solution into Equation

We know that \( x = \frac{5}{3} \) is a solution to the equation \( ax^2 + bx = 0 \). Substitute \( x = \frac{5}{3} \) into the equation to get: \[ a\left(\frac{5}{3}\right)^2 + b\left(\frac{5}{3}\right) = 0. \]

Step 2: Simplify the Expression

Calculate \( \left(\frac{5}{3}\right)^2 \): \[ \left(\frac{5}{3}\right)^2 = \frac{25}{9}. \] Now substitute it back into the equation:\[ a\left(\frac{25}{9}\right) + b\left(\frac{5}{3}\right) = 0. \]

Step 3: Clear the Fractions

To eliminate the fractions, multiply the entire equation by 9:\[ a \cdot 25 + b \cdot 15 = 0. \]

Step 4: Formulate Equation for Coefficients

From the equation in Step 3, we have:\[ 25a + 15b = 0. \]This equation represents the relationship between \( a \) and \( b \).

Step 5: Isolate One Variable

Solve for one variable in terms of the other. Let's solve for \( b \) in terms of \( a \): \[ b = -\frac{25}{15}a = -\frac{5}{3}a. \]

Step 6: Choose Specific Values for Simplification

Choose some value for \( a \) to determine \( b \). If \( a = 3 \), then:\[ b = -\frac{5}{3} \times 3 = -5. \] Thus, one possibility is \( a = 3 \) and \( b = -5 \).

Step 7: Verify the Solution

Substitute \( a = 3 \) and \( b = -5 \) back into the original equation to verify the solution:Translate back to the original form: \[ 3x^2 - 5x = 0 \]Substitute \( x = \frac{5}{3} \) and verify:\[ 3\left(\frac{5}{3}\right)^2 - 5\left(\frac{5}{3}\right) = 3\left(\frac{25}{9}\right) - 5\left(\frac{5}{3}\right) = 0. \]The left side reduces to zero, confirming the solution is correct.

Key Concepts

Solution Verification

When solving equations, it's important to ensure that the solutions we find actually work in the original equation. This process is called "solution verification." After determining potential solutions, we can check them by substituting the values back into the original equation.
Let's break this down with our exercise. Once we found the values of \( a = 3 \) and \( b = -5 \), we substituted these into the quadratic equation:

• The original equation is \( ax^2 + bx = 0 \).
• Substituting the values, we get \( 3x^2 - 5x = 0 \).

To verify, we plugged in the solution \( x = \frac{5}{3} \):

• Substituting \( x \), the equation becomes \( 3\left(\frac{5}{3}\right)^2 - 5\left(\frac{5}{3}\right) = 0 \).
• This simplifies to \( 0 = 0 \), confirming the solution is indeed correct.

This step is crucial to avoid errors and ensure the integrity of our solution. Solution verification should always be the final checkpoint in solving equations.

Substitution

Substitution is a key technique used to solve equations by replacing variables with known values. This helps in simplifying and solving the equations step by step.
In our scenario, we know \( x = \frac{5}{3} \) is a solution to the equation, so we substitute this value into the equation.
Let's look at how substitution is used in our exercise:

• Original equation: \( ax^2 + bx = 0 \).
• Substitute \( x = \frac{5}{3} \) into the equation: \( a\left(\frac{5}{3}\right)^2 + b\left(\frac{5}{3}\right) = 0 \).

This substitution converts the problem from one with a variable to a numerical expression, allowing us to conduct further simplifications. Substitution is particularly valuable when dealing with complex equations, as it reduces variables and opens up the way to solve them directly.

Variables Isolation

Isolating variables is a fundamental algebraic skill used to solve equations. The goal here is to express one variable in terms of another, making it easier to solve or analyze the equation.
In our exercise, we reached the equation \( 25a + 15b = 0 \) after clearing fractions. To isolate one variable, we express \( b \) in terms of \( a \):

• Rearrange the equation: \( 15b = -25a \).
• Divide by 15 on both sides: \( b = -\frac{25}{15}a \).
• Further simplifying gives us \( b = -\frac{5}{3}a \).

This transformation simplifies the relationship between \( a \) and \( b \). Isolating variables is powerful for solving equations, allowing you to choose specific values to further simplify or solve for exact solutions.

Fractions Elimination

Eliminating fractions from equations helps simplify calculations, making equations more manageable. This technique usually involves multiplying through by a common denominator to clear the fractions entirely.
In our exercise, after substitution, we ended up with the equation:

• \( a\left(\frac{25}{9}\right) + b\left(\frac{5}{3}\right) = 0 \).

To eliminate fractions:

• Identify the common denominator, which is 9.
• Multiply the entire equation by 9 to clear fractions: \( a \cdot 25 + b \cdot 15 = 0 \).

By removing fractions, equations become simpler to handle, allowing easier progression through further calculations. This step often precedes variable isolation, providing clearer expressions for manipulation.Removing fractions not only makes arithmetic simpler but also reduces opportunities for error in both calculation and interpretation.