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Magazine Chapter 1: Problem 22
Find all real solutions of the equation by completing the square. $$x^{2}-5 x+1=0$$
Short Answer
Expert verified
Solutions are \( x = \frac{5 + \sqrt{21}}{2} \) and \( x = \frac{5 - \sqrt{21}}{2} \).
Step by step solution
01
Move Constant to the Right Side
Start by rewriting the equation with the constant on the right side. The original equation is \( x^2 - 5x + 1 = 0 \). Subtract 1 from both sides to obtain: \[ x^2 - 5x = -1. \]
02
Prepare to Complete the Square
To complete the square, take the coefficient of \( x \), which is -5, divide it by 2, and square it. This is \( \left(\frac{-5}{2}\right)^2 = \frac{25}{4} \).
03
Add and Subtract the Square Term
Add and subtract \( \frac{25}{4} \) inside the left side of the equation to preserve the equality. \[ x^2 - 5x + \frac{25}{4} - \frac{25}{4} = -1. \] The equation becomes: \[ (x^2 - 5x + \frac{25}{4}) = -1 + \frac{25}{4}. \]
04
Simplify the Equation
Simplify the terms: \(-1 + \frac{25}{4} = \frac{-4}{4} + \frac{25}{4} = \frac{21}{4} \). The equation is now \( x^2 - 5x + \frac{25}{4} = \frac{21}{4} \).
05
Write Left Side as a Perfect Square
The expression \( x^2 - 5x + \frac{25}{4} \) can be written as a square of a binomial: \( (x - \frac{5}{2})^2 \). Thus, the equation is: \[ (x - \frac{5}{2})^2 = \frac{21}{4}. \]
06
Take the Square Root of Both Sides
Take the square root of both sides to solve for \( x \). \[ x - \frac{5}{2} = \pm \sqrt{\frac{21}{4}}. \] Simplifying the right side yields \( \pm \frac{\sqrt{21}}{2} \).
07
Solve for x
Add \( \frac{5}{2} \) to both sides to solve for \( x \): \[ x = \frac{5}{2} \pm \frac{\sqrt{21}}{2}. \] This gives the solutions: \[ x = \frac{5 + \sqrt{21}}{2} \] and \[ x = \frac{5 - \sqrt{21}}{2}. \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equations
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). Such equations graph as parabolas in a Cartesian coordinate plane. The solutions to these equations can be real or complex numbers.
Depending on the value of the discriminant \( b^2 - 4ac \), a quadratic equation can have:
Depending on the value of the discriminant \( b^2 - 4ac \), a quadratic equation can have:
- Two distinct real solutions, if the discriminant is positive.
- One real solution (also known as a repeated or double root), if the discriminant is zero.
- Two complex conjugate solutions, if the discriminant is negative.
Algebraic Methods
Completing the square is an essential algebraic technique typically used to solve quadratic equations. This method involves rewriting a quadratic expression so that it becomes a perfect square trinomial. Let's unpack how to complete the square using this method.
- Step 1: Isolate the quadratic and linear terms. This means moving the constant term to the right side of the equation as seen in \( x^2 - 5x = -1 \).
- Step 2: Find the magic number. To complete the square, take the coefficient of the linear term (which is \(-5\) here), halve it, and then square the result. So, \( \left( \frac{-5}{2} \right)^2 = \frac{25}{4} \) becomes the number to add and subtract.
- Step 3: Form a complete square. Add and subtract \( \frac{25}{4} \) on the left side: \( x^2 - 5x + \frac{25}{4} - \frac{25}{4} = -1 \). Rearrange it to \( \left(x - \frac{5}{2}\right)^2 = -1 + \frac{25}{4} \).
- Step 4: Simplify the equation. This step involves solving the equation \( \left(x - \frac{5}{2}\right)^2 = \frac{21}{4} \) by isolating \( x \). By taking the square root of both sides and solving for \( x \), you arrive at the solutions \( x = \frac{5 \pm \sqrt{21}}{2} \).
Problem Solving Steps
Understanding the step-by-step process of completing the square can greatly enhance one's problem-solving skills in algebra. This technique not only assists in reaching a solution but also provides a deeper insight into how quadratic expressions can be manipulated.
Each step adds vital information:
Each step adds vital information:
- Reorganizing the equation: Moving the constant to the opposite side clears the way for completing the term into a square.
- Calculating the necessary square: Determining the correct square term is crucial to forming a perfect square trinomial.
- Simplifying and solving: By reducing the complex equation into a simple binomial square, we see how effectively algebra can simplify seemingly complicated problems into a straightforward calculation.
