Problem 6 - Algebra 2

Question

Find the exact solutions by using the Quadratic Formula. \(x^{2}-2 x-2=0\)

Step-by-Step Solution

Answer

The exact solutions are \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).

Step 1: Identify Coefficients

The quadratic equation is in the form \( ax^{2} + bx + c = 0 \). In this equation, \( a = 1 \), \( b = -2 \), and \( c = -2 \).

Step 2: Write the Quadratic Formula

The Quadratic Formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula will be used to find the solutions of the given quadratic equation.

Step 3: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). Here, \( b^2 = (-2)^2 = 4 \) and \( 4ac = 4 \times 1 \times (-2) = -8 \). Thus, the discriminant is \( 4 + 8 = 12 \).

Step 4: Substitute into the Quadratic Formula

Use the values of \( a \), \( b \), and the discriminant in the Quadratic Formula. The formula becomes: \[ x = \frac{-(-2) \pm \sqrt{12}}{2 \times 1} \].

Step 5: Simplify the Solution

Simplify \( \frac{2 \pm \sqrt{12}}{2} \). \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \). Thus, the equation becomes \[ x = \frac{2 \pm 2\sqrt{3}}{2} \].

Step 6: Final Answer

Divide both terms inside the fraction by 2: \[ x = 1 \pm \sqrt{3} \]. So, the solutions are \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).

Key Concepts

Discriminant

The discriminant is a key component of the quadratic formula. It is found inside the square root portion of the formula and is denoted as \( b^2 - 4ac \). The discriminant helps to determine the nature of the roots of a quadratic equation.

• If the discriminant is positive, as in our example with a value of 12, the quadratic equation will have two distinct real solutions.
• If the discriminant is zero, the equation has exactly one real solution, meaning both roots are the same, also known as a repeated root.
• If the discriminant is negative, the solutions are complex or imaginary numbers, as the square root of a negative number is not a real number.

In our problem, the positive discriminant (12) indicates that there are two real and distinct solutions for the quadratic equation. Always calculate the discriminant first when using the quadratic formula, as it informs the subsequent steps of finding exact solutions.

Quadratic Equation

A quadratic equation is a polynomial equation of the second degree, which means it includes a variable raised to the power of two. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
These equations are called 'quadratic' because the highest power of the variable (\( x \)) is 2, which forms the shape of a parabolic curve when graphed.
To find the roots or solutions of a quadratic equation, we often use the Quadratic Formula, which is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula allows us to find the solutions for any quadratic equation by substituting the appropriate values of \( a \), \( b \), and \( c \). The quadratic formula is always applicable; however, remember that the discriminant's sign will affect the nature of the solutions.

Exact Solutions

When we solve a quadratic equation precisely without approximations, we refer to the results as exact solutions. The quadratic formula provides exact solutions in terms of radicals rather than decimal approximations.

• In our example, substituting into the quadratic formula and simplifying the expression \( \frac{2 \pm 2\sqrt{3}}{2} \) gives us the exact solutions \( x = 1 + \sqrt{3} \) and \( x = 1 - \sqrt{3} \).
• These solutions are exact because they preserve the radical form, \( \sqrt{3} \), instead of converting it into a decimal, which would result in an approximation.

Exact solutions are important in mathematics because they maintain the precision necessary in various theoretical applications. Using exact forms ensures that we do not lose any mathematical information that can occur with rounding or approximated decimals.