Problem 61 - Algebra for College Students

Question

Expressing solutions to the nearest one-thousandth. $$5 x^{2}-9 x+1=0$$

Step-by-Step Solution

Answer

The solutions to the equation are approximately \( x = 1.681 \) and \( x = 0.119 \).

Step 1: Identify the equation type

The equation given is a quadratic equation, which is in the form of \( ax^2 + bx + c = 0 \). Identifying that this is a quadratic equation helps in understanding that we can use the quadratic formula to find its roots.

Step 2: Write down the quadratic formula

Quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula will help us find the values of \(x\), by substituting values for \(a\), \(b\), and \(c\).

Step 3: Assign values for a, b, and c

For the equation \( 5x^2 - 9x + 1 = 0 \), assign \( a = 5 \), \( b = -9 \), and \( c = 1 \). These are the coefficients and the constant term of the quadratic equation.

Step 4: Calculate the discriminant

Calculate \( b^2 - 4ac \) using \( b = -9 \), \( a = 5 \), and \( c = 1 \). This becomes \( (-9)^2 - 4 \times 5 \times 1 = 81 - 20 = 61 \).

Step 5: Substitute into the quadratic formula

Substitute \( a \), \( b \), and the discriminant into the quadratic formula: \( x = \frac{-(-9) \pm \sqrt{61}}{2 \times 5} \). This simplifies to \( x = \frac{9 \pm \sqrt{61}}{10} \).

Step 6: Calculate the square root of the discriminant

Calculate \( \sqrt{61} \), which is approximately \( 7.810 \) to three decimal places. This can be done using a calculator.

Step 7: Find the two solutions

Calculate the two possible values for \( x \):1. \( x = \frac{9 + 7.810}{10} \rightarrow x = \frac{16.810}{10} \rightarrow x \approx 1.681 \)2. \( x = \frac{9 - 7.810}{10} \rightarrow x = \frac{1.190}{10} \rightarrow x \approx 0.119 \).

Step 8: Round to the nearest thousandth

Both solutions from the previous step are already rounded to three decimal places as required. Confirm that no further rounding is necessary.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. These are equations of the form \( ax^2 + bx + c = 0 \). The formula itself is expressed as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This means that, to find the values of \( x \), we need to know the coefficients \( a \), \( b \), and \( c \) which come from the equation. After finding these values, we substitute them into the formula. This formula will give us two possible solutions for \( x \), because of the "\( \pm \)" symbol. This leads us to the next concept, the discriminant, which helps determine the nature of these solutions.
Always remember to carefully assign the correct numerical values to \( a \), \( b \), and \( c \), and follow the order of operations when substituting them into the formula.

Discriminant

The discriminant is a vital part of the quadratic formula and plays a crucial role in determining the nature of the roots of a quadratic equation. The discriminant is calculated as \( b^2 - 4ac \). Here’s what it tells you:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (also called a repeated root).
• If it is negative, there are no real roots, but two complex roots.

In the equation \( 5x^2 - 9x + 1 = 0 \), the discriminant is \( 61 \), which is positive. This indicates that there are two distinct real roots. The positive discriminant ensures that the square root term in the quadratic formula \( \sqrt{b^2 - 4ac} \) is real and non-imaginary, critical for evaluating the solutions.

Roots of Quadratic Equation

The solutions or roots of a quadratic equation are the values of \( x \) that satisfy the equation. They can be found using the quadratic formula. In our example, substituting \( a = 5 \), \( b = -9 \), and \( c = 1 \) into the quadratic formula gives us:

• \( x = \frac{9 \pm \sqrt{61}}{10} \)

Because of the "\( \pm \)" sign, this gives two solutions:

• \( x_1 = \frac{9 + \sqrt{61}}{10} \)
• \( x_2 = \frac{9 - \sqrt{61}}{10} \)

These are the roots of the equation, and they represent the points where the parabola defined by the quadratic function crosses the x-axis. Calculating these roots may often involve using a calculator to find the decimal form of the solutions. The nature of these roots, such as whether they are real or complex, is determined by the discriminant.

Rounding to Thousandth

Rounding numbers to a given decimal place is often necessary in mathematics to simplify the answer for practical use. When rounding to the nearest thousandth, we focus on three places after the decimal point. This is essential in ensuring solutions remain accurate but also manageable in size.Here’s a quick guide:

• Look at the digit in the fourth decimal place.
• If this digit is 5 or more, round up the third decimal place by one.
• If it is less than 5, leave the third decimal place as it is.

For example, in finding the roots of the equation \( 5x^2 - 9x + 1 = 0 \), the solutions were \( x_1 \approx 1.681 \) and \( x_2 \approx 0.119 \). Each was already rounded to three decimal places, thus conforming to the requirement of rounding to the nearest thousandth. Rounding accurately helps in achieving precise results in computations and interpretations.