Problem 53 - Algebra 1

Question

Choose a method to solve the quadratic equation. Explain your choice. $$9 a^{2}-25=0$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation given are \(a = -5/3\) and \(a = 5/3\).

Step 1: Simplify The Equation

The equation \(9a^{2}-25=0\) can be written in its difference of squares form, as \((3a+5)(3a-5)= 0\).

Step 2: Equate Each Factor to Zero

Set each factor to equal zero, this gives two equations: \(3a+5= 0\) and \(3a -5=0\).

Step 3: Solve for the Variable

Solving each equation will gives the two roots of the quadratic equation. For \(3a + 5 = 0\), isolating \(a\) gives \(a = -5/3\). Similarly, for \(3a - 5 = 0\), isolating \(a\) gives \(a = 5/3\).

Key Concepts

Difference of Squares

Quadratic equations can often be solved by recognizing them as a "difference of squares." This method is applicable when a quadratic equation is structured as a square of a term minus another square equals zero. In our problem, we see:
- The equation is given as \(9a^2 - 25 = 0\).- This equation can be rewritten using the difference of squares identity: When you identify an expression in the form \(A^2 - B^2\), it can be factored as \((A + B)(A - B = 0)\). Here, \(9a^2\) is \((3a)^2\) and \(25\) is \(5^2\).
Thus, applying this transformation allows us to express:

• \(9a^2 - 25 = (3a+5)(3a-5) = 0\)

Recognizing the difference of squares is crucial as it simplifies the quadratic equation into a factorable form, making it much easier to solve.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as a product of simpler expressions set to zero. In our problem, after identifying the difference of squares, we factored the expression:

• \((3a + 5)(3a - 5) = 0\)

Factoring transforms the quadratic expression into two linear factors. The rationale behind factoring is that if the product of two factors is zero, then at least one of these factors must be zero. This principle is known as the "zero-product property."
Applying this property, we set each factor to zero and solve for the variable:

• \(3a + 5 = 0\)
• \(3a - 5 = 0\)

Factoring, coupled with the zero-product property, simplifies finding the roots of a quadratic equation, making it much more manageable especially with the difference of squares.

Roots of Equations

Finding the roots of an equation involves solving for values that satisfy the equation. In the context of quadratic equations, the roots are the values of the variable which, when substituted, make the equation equal to zero. After factoring the quadratic equation \(9a^2-25=0\), we obtain:

• \((3a + 5)(3a - 5) = 0\)

Setting each factor equal to zero helps to find the roots of the equation:

• Solving \(3a + 5 = 0\) gives \(a = -\frac{5}{3}\).
• Solving \(3a - 5 = 0\) gives \(a = \frac{5}{3}\).

These roots \(a = -\frac{5}{3}\) and \(a = \frac{5}{3}\) are where the graph of \(y = 9a^2 - 25\) intersects the x-axis. Understanding how to find roots is essential in solving quadratic equations since they represent the solutions. Each root shows where the function equals zero, providing important insights into the behavior of the graphed equation.