Problem 55 - Algebra and Trigonometry

Question

Solve equation by completing the square. $$ x^{2}-5 x+6=0 $$

Step-by-Step Solution

Answer

The solutions for the equation are \(x = 2\) and \(x = 3\).

Step 1: Identify a, b and c from the equation

The quadratic equation given is \(x^{2}-5 x+6=0\). Thus, the values are a = 1, b = -5, and c = 6

Step 2: Applying the completing square method

Rearrange the equation as \((x - \frac{b}{2a})^2 = \frac{b^2}{4a^2} - \frac{c}{a}\). That becomes \((x - \frac{-5}{2})^2 = \frac{(-5)^2}{4} - 6\), which simplifies to \((x + \frac{5}{2})^2 = 6.25 - 6\)

Step 3: Simplify the right hand side

Solving the right-hand side, we get \((x + \frac{5}{2})^2 = 0.25\)

Step 4: Take the square root of both sides

Finding the square root on both sides, we get \(x + \frac{5}{2} = ± \frac{1}{2}\)

Step 5: Solve for x

Finally, solving for x, we get two solutions: \(x_1 = \frac{5}{2} - \frac{1}{2} = 2\), and \(x_2 = \frac{5}{2} + \frac{1}{2} = 3\)

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree 2. It generally takes the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are coefficients and \(x\) represents the variable. Quadratics are quite common in algebra and appear in various real-world applications, including physics and engineering.
The significance of the quadratic equation lies in its symmetrical properties and its shape, the parabola.

• When the coefficient \(a > 0\), the parabola opens upwards.
• If \(a < 0\), it opens downwards.

The roots (or solutions) of a quadratic equation can be found using different methods like factoring, completing the square, or using the quadratic formula. Each method provides a unique way to solve the equation and find the points where the parabola intersects the x-axis.

Factoring

Factoring is a method used to solve quadratic equations by expressing them as a product of linear factors. For the equation \(x^2 - 5x + 6 = 0\), you would look for two numbers that multiply to \(6\) (the constant term, \(c\)) and add to \(-5\) (the coefficient of \(x\), \(b\)).
By observation, these numbers are \(-2\) and \(-3\), since \((-2)\times (-3) = 6\) and \(-2) + (-3) = -5\).
Thus, the quadratic equation can be rewritten and solved as:

• \((x - 2)(x - 3) = 0\)
• Setting each factor to zero gives: \(x - 2 = 0\) or \(x - 3 = 0\)
• Solving these gives us the solutions: \(x = 2\) and \(x = 3\)

Factoring is efficient when the quadratic can be easily decomposed into integers, which align well with the sum and product of the coefficients.

Square Root

The square root is a mathematical concept used to derive a number \(y\) such that \(y^2 = x\). Solving equations often involves taking square roots, especially in methods like completing the square.
Applying the square root method to the completed square form, \((x + \frac{5}{2})^2 = 0.25\), involves taking the square root of both sides.

• On the left-hand side, take the square root of \((x + \frac{5}{2})^2\), leaving us with \(x + \frac{5}{2}\).
• The square root of 0.25 is \(\pm 0.5\), giving the equations \(x + \frac{5}{2} = 0.5\) and \(x + \frac{5}{2} = -0.5\).

Taking the square root effectively helps simplify the quadratic to linear equations, which can be easily solved to find the roots.

Solving Equations

Solving equations involves finding the values of \(x\) that make the equation true. For quadratic equations, this often means finding the roots where the quadratic expression equals zero. There are several methods to achieve this, each with its trade-offs and conveniences.

• Completing the Square: This method restructures the equation into a form that allows taking square roots directly, providing precise solutions.
• Quadratic Formula: This universal method employs \(x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}\), which finds solutions for any quadratic equation.
• Graphing: Provides a visual approach, identifying where the parabola intersects the x-axis as roots.

Understanding these methods equips learners with various tools for tackling quadratic equations effectively, each suitable for different scenarios depending on the given data.