Problem 33 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. \(31-44\) : Solve by using the quadratic formula. $$ x^{2}+4 x+2=0 $$

Step-by-Step Solution

Answer

Solutions are \( x = -2 + \sqrt{2} \) and \( x = -2 - \sqrt{2} \).

Step 1: Identify a, b, c

The quadratic equation is in the form of \( ax^2 + bx + c = 0 \). Here, \( a = 1 \), \( b = 4 \), and \( c = 2 \).

Step 2: Write the Quadratic Formula

The quadratic formula is given by \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We'll need to substitute the values of \( a \), \( b \), and \( c \) into this formula.

Step 3: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \). For this equation, \( b^2 = 16 \) and \( 4ac = 8 \). So, the discriminant is \( 16 - 8 = 8 \).

Step 4: Substitute into the Quadratic Formula

Substitute \( b = 4 \), \( a = 1 \), and the calculated discriminant into the quadratic formula: \[ x = \frac{-4 \pm \sqrt{8}}{2} \]

Step 5: Simplify the Square Root of the Discriminant

Simplify \( \sqrt{8} \) as \( \sqrt{4 \times 2} = 2\sqrt{2} \). Now the formula becomes:\[ x = \frac{-4 \pm 2\sqrt{2}}{2} \]

Step 6: Simplify the Equation

Divide the numerator by the denominator:\[ x = \frac{-4}{2} \pm \frac{2\sqrt{2}}{2} \] This simplifies to:\[ x = -2 \pm \sqrt{2} \]

Step 7: Write the Solutions

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

Key Concepts

Discriminant

In the world of quadratic equations, the discriminant is a vital component. It determines the nature of the solutions that these equations may have. Found within the quadratic formula, the discriminant is the expression under the square root, represented as \( b^2 - 4ac \). For a given quadratic equation \( ax^2 + bx + c = 0 \), the value of the discriminant can provide insights into how many solutions exist and what kind they might be.

• If the discriminant is positive, meaning \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If it is zero, \( b^2 - 4ac = 0 \), there is exactly one real solution, known as a repeated or double root.
• If it is negative, \( b^2 - 4ac < 0 \), the equation has no real solutions, but instead two complex solutions.

In our specific problem, calculating the discriminant gives us the value \( 8 \), which is positive. Hence, we can conclude that the equation has two distinct real solutions.

Solutions of Quadratic Equations

Finding the solutions of a quadratic equation is like unveiling the roots where the curve intersects the x-axis. These solutions can be determined using multiple methods, with the quadratic formula being one of the most reliable. In general, a quadratic equation takes the form \( ax^2 + bx + c = 0 \), and the quadratic formula helps us find its solutions.
In our problem, after calculating the discriminant and employing the quadratic formula, we found the solutions to be \( x = -2 + \sqrt{2} \) and \( x = -2 - \sqrt{2} \). These solutions mean that the parabolic graph will cross the x-axis at these specific points.
The beauty of quadratic equations is that they have up to two solutions, indicated by the "\( \pm \)" symbol in the formula. Whether they are real, repeated, or complex solutions depends greatly on the discriminant value we calculate.

Quadratic Equation

A quadratic equation is a second-degree polynomial equation expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations are fundamental in algebra and appear frequently in various mathematical contexts.
Quadratic equations graph as parabolas, which can open upwards or downwards depending on the sign of \( a \). Their vertex, axis of symmetry, and x-intercepts (roots) are all features derived from the coefficients and values in the equation.
In our example equation, \( x^2 + 4x + 2 = 0 \), we identified the coefficients as \( a = 1 \), \( b = 4 \), and \( c = 2 \). These give us key parts of the equation needed to find its roots using the quadratic formula. Understanding each term's role helps simplify solving these types of problems while appreciating the elegance behind the mathematics of quadratic equations!