Problem 60 - Algebra for College Students

Question

Expressing solutions to the nearest one-thousandth. $$4 x^{2}-6 x+1=0$$

Step-by-Step Solution

Answer

The solutions are approximately 1.309 and 0.191.

Step 1: Identify the Type of Equation

The given equation is a quadratic equation of the form \( ax^2 + bx + c = 0 \) with \( a = 4 \), \( b = -6 \), and \( c = 1 \).

Step 2: Use the Quadratic Formula

The solutions for the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values of \( a \), \( b \), and \( c \) into the formula.

Step 3: Calculate the Discriminant

Compute the discriminant \( b^2 - 4ac \), which is \( (-6)^2 - 4 \times 4 \times 1 = 36 - 16 = 20 \).

Step 4: Substitute Values into the Formula

Substitute \( b = -6 \), \( a = 4 \), and the discriminant \( \sqrt{20} \) into the quadratic formula: \( x = \frac{6 \pm \sqrt{20}}{8} \).

Step 5: Simplify the Expression

The expression becomes \( x = \frac{6 \pm 2\sqrt{5}}{8} \). Simplify by dividing each term by 2, resulting in \( x = \frac{3 \pm \sqrt{5}}{4} \).

Step 6: Approximate and Express Solutions

Compute the approximate decimal values of \( \sqrt{5} \approx 2.236 \). This gives two solutions: \( x_1 = \frac{3 + 2.236}{4} \approx 1.309 \) and \( x_2 = \frac{3 - 2.236}{4} \approx 0.191 \).

Step 7: Round to the Nearest Thousandth

Round the solutions to the nearest one-thousandth: \( x_1 \approx 1.309 \) and \( x_2 \approx 0.191 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for finding the solutions to quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \). This formula is especially helpful when the equation does not factorize nicely. It gives solutions through the formula:

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

To use it effectively:

1. Identify the coefficients \( a \), \( b \), and \( c \) from the quadratic equation.
2. Substitute these values into the formula.
3. Solve for \( x \), which will typically yield two solutions because of the \( \pm \) symbol.

Discriminant

The discriminant is a critical part of the quadratic formula that helps determine the nature of the solutions without fully solving the equation. It is given by the expression \( b^2 - 4ac \). The value of the discriminant tells us:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution, a "double root."
• If \( b^2 - 4ac < 0 \), the solutions are complex or imaginary.

When solving \( 4x^2 - 6x + 1 = 0 \), we found the discriminant to be \( (-6)^2 - 4 \times 4 \times 1 = 20 \), indicating two distinct real solutions.

Solving Quadratic Equations

Solving quadratic equations involves finding the values of \( x \) that satisfy the equation, and we can solve them using a few methods. Here, we used the quadratic formula on \( 4x^2 - 6x + 1 = 0 \):

• Substitute \( a = 4 \), \( b = -6 \), and the discriminant \( \sqrt{20} \) into the formula
• This became \( x = \frac{6 \pm 2\sqrt{5}}{8} \)
• Then simplify by dividing each term by 2 to get \( x = \frac{3 \pm \sqrt{5}}{4} \)

This simplification process is essential for making the solutions easier to work with before approximating them to decimal forms.

Rounding Decimals

After simplifying the expressions, we often need to approximate the values to be more workable or meaningful, especially in practical scenarios. Here, it's crucial to round to the nearest thousandth – three decimal places:

• First, approximate \( \sqrt{5} \approx 2.236 \)
• Compute the decimal solutions: \( x_1 = \frac{3 + 2.236}{4} \approx 1.309 \) and \( x_2 = \frac{3 - 2.236}{4} \approx 0.191 \)
• Finally, ensure both solutions are rounded correctly to the nearest thousandth.

Rounding ensures that your solutions are as accurate and comprehensible as possible for real-world applications or further calculations.