Problem 44 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ 0=x^{2}-4 x+1 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \).

Step 1: Identify the Quadratic Equation

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

Step 2: Use the Quadratic Formula

The solution for a quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Substitute \( a = 1 \), \( b = -4 \), and \( c = 1 \) into the formula.

Step 3: Calculate the Discriminant

Calculate the discriminant, \( b^2 - 4ac \), which is part of the quadratic formula. For this equation, \[ b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 1 = 16 - 4 = 12 \]. Since the discriminant is positive, there will be two distinct real solutions.

Step 4: Compute the Solutions

Substitute the discriminant and the given coefficients back into the quadratic formula to find the solutions: \[ x = \frac{-(-4) \pm \sqrt{12}}{2 \cdot 1} = \frac{4 \pm \sqrt{12}}{2} \].

Step 5: Simplify the Solutions

Simplify \( \sqrt{12} \) as \( 2\sqrt{3} \), then simplify the entire expression to get the two solutions: \[ x = \frac{4 \pm 2\sqrt{3}}{2} \]. Simplifying further yields \[ x = 2 \pm \sqrt{3} \].

Step 6: Present the Real Solutions

The real solutions to the original equation are \( x = 2 + \sqrt{3} \) and \( x = 2 - \sqrt{3} \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that allows us to solve any quadratic equation, regardless of its complexity. A quadratic equation generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) represents the variable. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula directly provides the solutions for the variable \( x \), offering a straightforward way to find solutions when factoring is not easy or when the equation does not fit nicely into perfect squares.
When using this formula, remember:

• Identify the coefficients \( a \), \( b \), and \( c \) from your equation.
• Substitute these values into the formula carefully.

The quadratic formula is especially beneficial for equations that are not easily factorable, ensuring that solutions are found efficiently.

Discriminant

The discriminant is a key part of the quadratic formula that helps us understand the nature of the solutions, often without computation. It is the expression \( b^2 - 4ac \), found under the square root within the quadratic formula.
The value of the discriminant indicates:

• **Positive Discriminant:** The equation has two distinct real solutions, implying that the parabola defined by the quadratic equation crosses the x-axis at two points.
• **Zero Discriminant:** There is exactly one real solution, or a repeated root, meaning the parabola touches the x-axis at a single point.
• **Negative Discriminant:** There are no real solutions since the parabola does not intersect the x-axis; the solutions are complex or imaginary.

In our exercise, the calculation \((-4)^2 - 4 \cdot 1 \cdot 1 = 12\) results in a positive discriminant \(12\). Thus, the equation has two distinct real solutions.

Real Solutions

Once the discriminant and all required values are substituted into the quadratic formula, we can determine the real solutions of the given quadratic equation. Real solutions are the values of \( x \) that satisfy the equation, making it true.
In this task, using the quadratic formula with our values, \[ x = \frac{4 \pm \sqrt{12}}{2}, \] the simplification process leads to the expression \( x = \frac{4 \pm 2\sqrt{3}}{2} \).
After further simplification, we find:\[x = 2 + \sqrt{3} \quad \text{and} \quad x = 2 - \sqrt{3}.\]These are the two real solutions to our quadratic equation \( x^2 - 4x + 1 = 0 \). Remember, each step of calculation should be double-checked to ensure that the solutions align with the behavior noted from the discriminant. This accurate approach helps in clearly visualizing the points where the quadratic expression equals zero, illustrating the solutions effectively.