Problem 28 - College Algebra and Calculus: An Applied Approach

Question

Solve the quadratic equation by extracting square roots. List both the exact answer and a decimal answer that has been rounded to two decimal places. $$ 9 x^{2}=25 $$

Step-by-Step Solution

Answer

The solutions for the equation are exactly \( x = \pm 5/3 \) and approximately \( x = 1.67, -1.67 \)

Step 1: Isolate \( x^2 \)

To isolate \( x^2 \), divide both sides of the equation by 9. So, the updated equation is \( x^2 = 25/9 \).

Step 2: Extract the square roots

Take the square root of both sides. Remember to include the plus or minus sign before the square-root on the right-hand side because a number can have both positive and negative square roots. So, \( x = \pm \sqrt{25/9} \).

Step 3: Simplify the square root

This gives us \( x = \pm 5/3 \)

Step 4: Calculate decimal approximation

Lastly, convert the fractional values to decimal, rounded to two decimal places. For \( +5/3 \) it is approximately \( 1.67 \), for \( -5/3 \), it is approximately \( -1.67 \).

Key Concepts

Extracting Square Roots

When faced with a quadratic equation like \(9x^{2}=25\), one way to solve for \(x\) is by extracting square roots. This method involves several steps which lead to finding the exact values of \(x\). The process starts by isolating the \(x^2\) term on one side of the equation, which consequently makes the other side of the equation a perfect square. In this example, we start with dividing both sides by 9, yielding the simplified form \(x^2 = \frac{25}{9}\).

The next stage involves taking the square root of both sides. For an equation like this, there will always be two solutions—because both a positive and a negative number squared will give us the positive number we started with. This duality is represented by the 'plus-minus' symbol (\(\pm\)). When extracting the square roots here, it gives us \(x = \pm \sqrt{25/9}\), which simplifies to \(x = \pm \frac{5}{3}\).

Understanding the rationale behind including the plus and minus signs is crucial as it acknowledges both square roots of a number. By mastering this concept, students can solve a variety of quadratic equations effectively.

Quadratic Formula

The quadratic formula provides a fail-safe way to solve any quadratic equation, and it is especially useful when other methods, like factoring or completing the square, are not suitable. The formula is \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\), where \(a\), \(b\), and \(c\) are the coefficients of the terms in the standard form of a quadratic equation, \(ax^2 + bx + c = 0\).

For the equation \(9x^{2}=25\), we can also use the quadratic formula. However, since it is already a simple square and the other terms (\(b\) and \(c\)) are zero or absent, extracting square roots is more efficient. Nevertheless, understanding how to apply the quadratic formula in more complex cases is essential for students, as it ensures that they can tackle any quadratic problem they're presented with.

Decimal Approximation

While exact answers are often preferred in mathematics, many real-world applications require decimal approximations for practicality. When it comes to quadratic equations, once you've found the exact solutions, like \(x = \pm \frac{5}{3}\) from our example, you can convert these fractions into decimal form. This is particularly valuable when a clear decimal representation is necessary, such as in measurements or statistics.

To find a decimal approximation, you would simply divide the numerator by the denominator. In the case of \(\frac{5}{3}\), performing the division gives us a decimal value which can then be rounded to a specified degree of precision. For instance, when these values need to be reported to two decimal places, \(+\frac{5}{3}\) will be approximately 1.67 and \(-\frac{5}{3}\) approximately -1.67. It's important for students to practice this conversion to become comfortable with interpreting and reporting numerical data in its most applicable form.