Problem 35 - Elementary Algebra

Question

Use the quadratic formula to solve. $$ x 2-6 x+6=0 $$

Step-by-Step Solution

Answer

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

Step 1: Identify coefficients

The quadratic equation is given as \(x^2 - 6x + 6 = 0\). First, identify the coefficients \(a\), \(b\), and \(c\). In this equation, \(a = 1\), \(b = -6\), and \(c = 6\).

Step 2: Write the quadratic formula

The quadratic formula to solve for \(x\) in an equation \(ax^2 + bx + c = 0\) is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 3: Calculate the discriminant

Substitute the values of \(b\), \(a\), and \(c\) into the discriminant \(b^2 - 4ac\). Compute \((-6)^2 - 4(1)(6) = 36 - 24 = 12\). The discriminant is 12.

Step 4: Solve using the quadratic formula

Substitute \(b = -6\), \(a = 1\), and the discriminant value into the quadratic formula: \(x = \frac{-(-6) \pm \sqrt{12}}{2(1)}\), which simplifies to \(x = \frac{6 \pm \sqrt{12}}{2}\).

Step 5: Simplify the expression

Further simplify the expression: \(x = \frac{6 \pm \sqrt{4 \cdot 3}}{2}\). Since \(\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}\), substitute back to get \(x = \frac{6 \pm 2\sqrt{3}}{2}\).

Step 6: Final simplification

Divide each term inside the fraction by 2 to simplify: \(x = 3 \pm \sqrt{3}\). The solutions to the equation are \(x = 3 + \sqrt{3}\) and \(x = 3 - \sqrt{3}\).

Key Concepts

Discriminant

The discriminant is a key part of the quadratic formula and helps determine the nature of the roots of a quadratic equation. It is found within the quadratic formula under the square root: \(b^2 - 4ac\). Let's break this down:

• If the discriminant, \(b^2 - 4ac\), is greater than zero, the quadratic equation has two distinct real solutions. This means the parabola intersects the x-axis at two points.
• If the discriminant equals zero, there is exactly one real solution, or we can say the root is repeated. In this case, the vertex of the parabola just touches the x-axis.
• If the discriminant is less than zero, the equation has no real solutions, implying the parabola does not intersect the x-axis, and the solutions are complex.

In our example, the discriminant is \(12\), which is positive. Thus, the quadratic equation \(x^2 - 6x + 6 = 0\) has two distinct real roots.

Quadratic Equation

Quadratic equations are fundamental expressions in algebra and take the general form \(ax^2 + bx + c = 0\). These equations are called "quadratic" because the highest exponent of the variable is 2. Here's a bit more detail:

• The coefficient \(a\) cannot be zero because if it were, the equation would no longer be quadratic but linear instead.
• The term \(b\) is part of the expression that involves the first power of \(x\).
• Finally, \(c\) is the constant term and does not include \(x\) at all.

To solve these equations, we often use the quadratic formula, which is particularly handy when factoring is not obvious or easy. In this exercise, the quadratic equation provided is \(x^2 - 6x + 6 = 0\), where \(a = 1\), \(b = -6\), and \(c = 6\). By applying the quadratic formula, you can find the roots of the equation efficiently.

Solution Steps

Solving a quadratic equation with the quadratic formula involves a systematic approach, which ensures accuracy:1. **Identify Coefficients** - From the equation \(x^2 - 6x + 6 = 0\), coefficients are \(a = 1\), \(b = -6\), \(c = 6\).2. **Apply the Quadratic Formula** - The formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Plug in the known values for a clean start.3. **Calculate the Discriminant** - Calculate \(b^2 - 4ac = 12\). A positive discriminant indicates two real solutions.4. **Substitute Back Into The Formula** - Replace \(b\), \(a\), and the discriminant value into the formula: \(x = \frac{-(-6) \pm \sqrt{12}}{2(1)}\).5. **Simplify the Expressions** - Simplify \(x = \frac{6 \pm \sqrt{12}}{2}\) further into \(x = \frac{6 \pm 2\sqrt{3}}{2}\).6. **Final Simplification** - Divide through to obtain: \(x = 3 \pm \sqrt{3}\).These steps guide you from identifying components to simplifying for the final answer, making the quadratic formula a powerful tool for finding solutions effectively.