Problem 33 - Algebra 2

Question

Solve each equation by completing the square. \(2 x^{2}-3 x+1=0\)

Step-by-Step Solution

Answer

The solutions are \(x = 1\) and \(x = \frac{1}{2}\).

Step 1: Adjust the Equation

First, adjust the equation by dividing all terms by 2 to make the coefficient of \(x^2\) equal to 1. This gives: \(x^2 - \frac{3}{2}x + \frac{1}{2} = 0\).

Step 2: Move Constant Term

Move the constant term to the other side of the equation to focus on the quadratic and linear terms: \(x^2 - \frac{3}{2}x = -\frac{1}{2}\).

Step 3: Calculate the Square of Half the Coefficient of x

Take the coefficient of \(x\), which is \(-\frac{3}{2}\), divide it by 2 to get \(-\frac{3}{4}\), and then square it to get \(\left(-\frac{3}{4}\right)^2 = \frac{9}{16}\).

Step 4: Complete the Square

Add \(\frac{9}{16}\) to both sides of the equation to balance it, resulting in: \[x^2 - \frac{3}{2}x + \frac{9}{16} = -\frac{1}{2} + \frac{9}{16}\].

Step 5: Simplify the Equation

Simplify the right side: \[-\frac{1}{2} + \frac{9}{16} = -\frac{8}{16} + \frac{9}{16} = \frac{1}{16}\].Now the equation is: \[x^2 - \frac{3}{2}x + \frac{9}{16} = \frac{1}{16}\].

Step 6: Express as a Perfect Square

The left side of the equation now forms a perfect square: \[(x - \frac{3}{4})^2 = \frac{1}{16}\].

Step 7: Solve for x Using Square Roots

Take the square root of both sides, remembering the \(\pm\) sign: \[x - \frac{3}{4} = \pm \frac{1}{4}\].

Step 8: Finalize the Solution for x

Solve for \(x\) by adding \(\frac{3}{4}\) to both sides for each case: 1. \(x - \frac{3}{4} = \frac{1}{4}\) gives \(x = \frac{3}{4} + \frac{1}{4} = 1\).2. \(x - \frac{3}{4} = -\frac{1}{4}\) gives \(x = \frac{3}{4} - \frac{1}{4} = \frac{1}{2}\).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental part of algebra. They have many applications, from physics to finance. A standard form of a quadratic equation is expressed as:
\[ ax^2 + bx + c = 0 \]

• Here, \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable.
• The value of \( a \) cannot be zero, as this would not form a quadratic equation anymore.
• The most common methods to solve these equations include factoring, using the quadratic formula, and completing the square.

Completing the square is a method used to convert a quadratic equation into a perfect square trinomial:
\[ (x + p)^2 = q \]This method paves the way for solving equations using the square root method.

Square Root Method

The square root method is key in solving quadratic equations once they are expressed as a perfect square. This usually involves taking the square root of both sides of the equation:
\[ (x - h)^2 = k \]Here’s how it works:

• First, identify the perfect square trinomial on the left side of the equation.
• Once expressed in the form \((x - h)^2\), take the square root of both sides. This results in two possible solutions because of the nature of square roots:

Taking the square root introduces a \(\pm\) symbol:
\[ x - h = \pm \sqrt{k} \]This step generates two potential values for \( x \).
Finally, solve for \( x \) by isolating the variable, leading to two solutions. This method is straightforward but requires care to ensure accuracy, especially when dealing with fractions.

Polynomial Equations

Polynomial equations, like quadratics, consist of variables with whole number exponents.
They include terms such as \( x^n \), where \( n \) is an integer:

• Polynomial equations of degree 2 are quadratics, characterized by the term \( x^2 \).
• Larger degree polynomials include cubic (degree 3) and quartic (degree 4) equations.

For solving polynomial equations, several strategies are available:

• Using algebraic techniques such as factoring or synthetic division.
• Applying numerical methods or graphing for visualization.
• The roots or solutions of polynomial equations represent where the polynomial is equal to zero.

Understanding the nature and solutions of polynomials is crucial to mastering calculus and advanced algebra. Quadratic equations play a significant role within polynomial concepts as they are the simplest form beyond linear equations, making them foundational for further studies.