Problem 75 - Algebra 1: Concepts and Skills

Question

Use the quadratic formula to solve the equation. If the solution involves radicals, round to the nearest hundredth. $$4 x^{2}-6 x+1=0$$

Step-by-Step Solution

Answer

The solutions to the equation \(4x^2 - 6x + 1 = 0\) are \(x = 1.29\) and \(x = 0.21\), when rounded to the nearest hundredth.

Step 1: Identify values for a, b, and c

We have the quadratic equation in the form \(ax^2 + bx + c = 0\). Here, \(a = 4\), \(b = -6\), and \(c = 1\). We will use these values in the quadratic formula.

Step 2: Apply the Quadratic Formula

The Quadratic Formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the values of \(a\), \(b\), and \(c\) into this formula. Thus, \(x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4*4*1}}{2*4}\).

Step 3: Solve for x

Simplify the expression under the square root and the fraction to find the values of x. Calculate \(x = \frac{6 \pm \sqrt{36 - 16}}{8}\), then it simplifies to \(x = \frac{6 \pm \sqrt{20}}{8}\). The solutions in decimal form, rounded to two decimal places, would be \(x1 = 1.29\) and \(x2 = 0.21\).

Key Concepts

Understanding Quadratic Equations

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). The term \( ax^2 \) is what makes the equation quadratic. These equations often appear in real-world problems involving areas, speeds, and other measurements.

Quadratic equations can have different types of solutions:

• Two real solutions
• One real solution
• No real solutions (when solutions are complex numbers)

To solve a quadratic equation, you can use methods like factoring, completing the square, or the quadratic formula. The quadratic formula is particularly useful when the equation is difficult to factor.

Dealing with Radicals

When you solve quadratic equations using the quadratic formula, you often encounter square roots, also known as radicals. The expression \( \sqrt{b^2 - 4ac} \) in the quadratic formula can lead to radical numbers, especially if the discriminant \( b^2 - 4ac \) is not a perfect square.

Radicals can sometimes be simplified by looking for perfect square factors within them. For example, if you have \( \sqrt{20} \), you can simplify this to \( 2 \times \sqrt{5} \). However, in many cases, such as when dealing with decimal solutions, it's common to approximate radicals using a calculator to obtain a decimal answer.

Understanding how to work with radicals is crucial, as they are encountered frequently in higher mathematics and physics.

The Importance of Rounding Decimals

Rounding decimals is essential, especially when dealing with solutions that involve irrational numbers, like those containing radicals. When you round a decimal, you approximate it to a specific place value such as the nearest tenth, hundredth, or thousandth.

In many math problems, including solving quadratic equations, you may be asked to round your final answer. For example, if your calculation gives you \( 1.285714 \), you may need to round it to the nearest hundredth, which would result in \( 1.29 \).

Rounding helps present numbers in a simpler and more understandable form, making them easier to read and interpret in practical scenarios, like measurements in engineering and sciences.

Steps in Solving Equations

Solving equations, particularly quadratic ones, requires a systematic approach. Here’s how you could solve a quadratic equation like \( 4x^2 - 6x + 1 = 0 \) using the quadratic formula:

• Identify the coefficients \( a, b, \) and \( c \) from the equation.
• Substitute these into the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• Simplify under the square root to find the discriminant.
• Calculate the values of \( x \) using the plus and minus in the formula.
• Approximate the solutions if necessary, rounding to appropriate decimal places.

Each step is designed to progressively work towards isolating the variable \( x \). This approach allows you to systematically narrow down potential values of \( x \) and find the exact or approximate solutions.