Problem 40 - Algebra and Trigonometry

Question

Find all real solutions of the equation. $$ x^{2}-6 x+1=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Equation Form

The equation given is a quadratic equation in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -6 \), and \( c = 1 \).

Step 2: Calculate the Discriminant

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). Substitute the values to find the discriminant:\[\Delta = (-6)^2 - 4 \times 1 \times 1 = 36 - 4 = 32.\]The discriminant \( \Delta = 32 \) indicates that there are two distinct real roots.

Step 3: Use the Quadratic Formula

The quadratic formula for finding the roots of \( ax^2 + bx + c = 0 \) is:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]Substitute \( a = 1 \), \( b = -6 \), and \( c = 1 \) into the formula:\[x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \times 1 \times 1}}{2 \times 1} = \frac{6 \pm \sqrt{32}}{2}.\]

Step 4: Simplify the Roots

First, simplify the square root term:\[ \sqrt{32} = \sqrt{16 \times 2} = 4\sqrt{2}.\]Now substitute back into the formula:\[x = \frac{6 \pm 4\sqrt{2}}{2} = \frac{6}{2} \pm \frac{4\sqrt{2}}{2} = 3 \pm 2\sqrt{2}.\]Therefore, the solutions are \( x = 3 + 2\sqrt{2} \) and \( x = 3 - 2\sqrt{2} \).

Key Concepts

Understanding the Discriminant

The discriminant is a crucial part of solving quadratic equations, giving insight into the nature of the roots. It is represented by the symbol \( \Delta \) and is calculated using the formula \( \Delta = b^2 - 4ac \).
For the quadratic equation \( ax^2 + bx + c = 0 \), the discriminant helps us determine the type of solutions the equation has.

• If \( \Delta > 0 \), there are two distinct real solutions, meaning two separate numbers can solve the equation.


• If \( \Delta = 0 \), there is exactly one real solution, which is also called a repeated or double root.


• If \( \Delta < 0 \), the equation has no real solutions, indicating that the solutions are complex numbers.

In our example, the discriminant is \( 32 \), which is greater than zero, indicating that the equation has two distinct real solutions.

Using the Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations, especially useful when factoring is impractical.
The formula is expressed as:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
This formula derives from completing the square method, and always works as long as you substitute the coefficients \( a \), \( b \), and \( c \) correctly.
To find the solutions of the quadratic equation \( x^2 - 6x + 1 = 0 \), we substitute:

• \( a = 1 \)


• \( b = -6 \)


• \( c = 1 \)

The formula then becomes:\[x = \frac{6 \pm \sqrt{32}}{2}\]
Calculating further gives us the simplified roots.

Interpreting Real Solutions

Real solutions of a quadratic equation refer to the x-values where the parabola defined by the equation crosses the x-axis.
They represent actual numbers on the number line, as opposed to complex solutions which include imaginary components.
With the quadratic formula, once the discriminant is determined to be greater than zero, such as in our example with \( \Delta = 32 \), it confirms two real solutions.
The roots \( x = 3 + 2\sqrt{2} \) and \( x = 3 - 2\sqrt{2} \) are real because the math yields exact numerical values when calculated.
This indicates that if you were to graph the equation \( x^2 - 6x + 1 = 0 \), the graph would intersect the x-axis at these two points.

• These points are the actual solutions to the equation.


• They provide crucial information about where the function equals zero.

Thus, finding real solutions helps us understand the behavior of quadratic functions in real-world applications.