Algebra Problem: Finding the Value of \(a^2 + \frac{16}{a^2}\)

We are given an algebraic equation \(a - \frac{12}{a} = 1\), where \(a > 0\). Our goal is to find the numerical value of the expression \(a^2 + \frac{16}{a^2}\).

Let's start with the given equation:

\(a - \frac{12}{a} = 1\)

To get rid of the fraction, we can multiply the entire equation by \(a\). Since we are given that \(a > 0\), multiplying by \(a\) will not change the direction of any inequality, and \(a\) is not zero, so it's a valid operation.

\(a \left(a - \frac{12}{a}\right) = 1 \cdot a\)

\(a^2 - a \cdot \frac{12}{a} = a\)

\(a^2 - 12 = a\)

Now, let's rearrange this equation into a standard quadratic form by moving all terms to one side:

\(a^2 - a - 12 = 0\)

This is a quadratic equation of the form \(Ax^2 + Bx + C = 0\), where \(A=1\), \(B=-1\), and \(C=-12\). We can solve this equation for \(a\) by factoring, completing the square, or using the quadratic formula. Factoring is often the simplest method if possible.

We need to find two numbers that multiply to -12 and add up to -1. These numbers are -4 and 3.

So, we can factor the quadratic equation as follows:

\((a - 4)(a + 3) = 0\)

For this product to be zero, at least one of the factors must be zero.

• Case 1: \(a - 4 = 0 \implies a = 4\)
• Case 2: \(a + 3 = 0 \implies a = -3\)

We have found two possible values for \(a\): 4 and -3.

However, the problem statement specifies that \(a > 0\). Therefore, we must choose the positive value of \(a\).

So, the value of \(a\) is 4.

Now that we have the value of \(a\), we can substitute it into the expression \(a^2 + \frac{16}{a^2}\) to find its value.

Substitute \(a = 4\) into the expression:

\(a^2 + \frac{16}{a^2} = (4)^2 + \frac{16}{(4)^2}\)

\(= 16 + \frac{16}{16}\)

\(= 16 + 1\)

\(= 17\)

Thus, the value of \(a^2 + \frac{16}{a^2}\) is 17.

Let's check this result against the given options.

The calculated value, 17, matches Option 3.

Revision Table: Key Steps in Solving the Algebra Problem

Additional Information: Solving Quadratic Equations

A quadratic equation is an equation of the form \(ax^2 + bx + c = 0\), where \(x\) is the variable and \(a\), \(b\), and \(c\) are constants with \(a \neq 0\). In our problem, the variable is \(a\).

Methods to solve quadratic equations include:

• Factoring: If the quadratic expression can be factored into two linear factors, say \((px + q)(rx + s) = 0\), then the solutions are \(x = -\frac{q}{p}\) and \(x = -\frac{s}{r}\). This was the method used in this solution.
• Completing the Square: This method involves manipulating the equation to form a perfect square trinomial on one side.
• Quadratic Formula: The solutions to \(ax^2 + bx + c = 0\) are given by the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). This formula can solve any quadratic equation.

In this specific problem, factoring \((a-4)(a+3)=0\) was straightforward because we could easily find integer factors of -12 that sum to -1.