Problem 68 - College Algebra Essentials

Question

Solve each equation in Exercises \(65-74\) using the quadratic formula. $$ x^{2}+5 x+2=0 $$

Step-by-Step Solution

Answer

The solutions to \( x^2 + 5x + 2 = 0 \) are \( x = -0.438 \) and \( x = -4.561 \) using decimal rounding to three digits.

Step 1: Identify the values of a, b, and c

In the given equation \( x^2 + 5x + 2 = 0 \), the equivalents are \( a = 1, b = 5, c = 2 \).

Step 2: Substitute the values in the Quadratic formula

We'll substitute these values in the quadratic formula to get \( x = \frac{-b ± sqrt(b^2 - 4ac)}{2a} = \frac{-5 ± sqrt((5)^2 - 4*(1)*(2))}{2*1} = \frac{-5 ± sqrt(25 - 8)}{2} = \frac{-5 ± sqrt(17)}{2} \).

Step 3: Calculate the roots

Now, simplifying the above calculation gives us the roots of the given equation. \( x_1 = \frac{-5 + sqrt(17)}{2} \approx -0.438 , x_2 = \frac{-5 - sqrt(17)}{2} \approx -4.561\)

Key Concepts

Understanding a Quadratic Equation

A quadratic equation is a type of polynomial equation of the second degree, which means it has the general form \( ax^2 + bx + c = 0 \). Here, \( a \), \( b \), and \( c \) are constants where \( a eq 0 \). This equation forms a parabola when graphed on a coordinate plane. The term \( ax^2 \) is the quadratic term, \( bx \) is the linear term, and \( c \) is the constant term. Quadratic equations are central in algebra because they describe the motion of objects under gravity, among other phenomena. They're solved using methods like factoring, completing the square, or the quadratic formula. For equations not easily factored or simplified by other methods, the quadratic formula is a reliable tool.

The Process of Solving Equations

Solving equations means finding the value(s) of the variable that makes the equation true. For quadratic equations, the main goal is to determine the values of \( x \) that satisfy it. In many cases, quadratic equations are solved by:

• Factoring the quadratic expression.
• Completing the square to transform it into a perfect square trinomial.
• Using the quadratic formula when the above methods aren't feasible.

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) offers a straightforward way to find solutions, particularly when equations are complex or cannot be easily factored.

Finding the Roots of a Quadratic

The roots of a quadratic equation are the values of \( x \) that satisfy the equality \( ax^2 + bx + c = 0 \). These roots can be real or complex numbers, and they are the points where the graph of the equation intersects the x-axis.For any quadratic equation, the number of real roots is determined by the discriminant, \( b^2 - 4ac \):

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is one real root (also known as a repeated or double root).
• If it is negative, the roots are complex (non-real) numbers.

In the exercise example, the discriminant \( 17 \) is positive, so the equation has two distinct real roots.

Step-by-Step Solution Using the Quadratic Formula

Using the quadratic formula involves identifying the coefficients \( a \), \( b \), and \( c \) from the quadratic equation. In the given equation, \( x^2 + 5x + 2 = 0 \), these values are \( a = 1 \), \( b = 5 \), and \( c = 2 \).Substitute these into the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}\]This simplifies to:\[x = \frac{-5 \pm \sqrt{17}}{2}\]From here, calculate the two potential solutions:

• \( x_1 = \frac{-5 + \sqrt{17}}{2} \approx -0.438 \)
• \( x_2 = \frac{-5 - \sqrt{17}}{2} \approx -4.561 \)

These values are the roots of the quadratic equation, showing where the graph intersects the x-axis.