Problem 35 - Contemporary Precalculus

Question

Use the quadratic formula to solve the equation. $$x^{2}+6 x+7=0$$

Step-by-Step Solution

Answer

Answer: The solutions are $$x_1 = \frac{-6 + \sqrt{8}}{2}$$ and $$x_2 = \frac{-6 - \sqrt{8}}{2}$$.

Step 1: Write down the equation and identify the coefficients

The given equation is $$x^{2}+6 x+7=0$$, which is a quadratic equation of the form $$ax^2+bx+c=0$$ where a = 1, b = 6, and c = 7.

Step 2: Apply the quadratic formula

Use the quadratic formula $$x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$$ to find the two possible solutions for x: Substitute a = 1, b = 6, and c = 7 into the quadratic formula: $$x = \frac{-6 \pm \sqrt{6^{2}-4 \cdot 1 \cdot 7}}{2 \cdot 1}$$

Step 3: Simplify the expression under the square root

Calculate the expression inside the square root: $$6^2 - 4 \cdot 1 \cdot 7 = 36 - 28$$ which results in 8. Now, our formula becomes: $$x = \frac{-6 \pm \sqrt{8}}{2}$$

Step 4: Solve for the two possible solutions

We now have two possible solutions for x: $$x_1 = \frac{-6 + \sqrt{8}}{2} \text{ and } x_2 = \frac{-6 - \sqrt{8}}{2}$$ So, the solutions to the quadratic equation $$x^{2}+6 x+7=0$$ are $$x_1 = \frac{-6 + \sqrt{8}}{2}$$ and $$x_2 = \frac{-6 - \sqrt{8}}{2}$$.

Key Concepts

Quadratic Equation

A quadratic equation is a polynomial equation of degree two. This means it has the highest exponent of 2. A general form of a quadratic equation is given by

• \( ax^2 + bx + c = 0 \)

where

• \( ax^2 \) is the quadratic term,
• \( bx \) is the linear term, and
• \( c \) is the constant term.

The goal is to find the value(s) of \( x \) that make this equation true. In our specific equation \( x^2 + 6x + 7 = 0 \), we see that

• the quadratic term has \( a = 1 \),
• the linear term has \( b = 6 \), and
• the constant term is \( c = 7 \).

These values \( a \), \( b \), and \( c \) will help us in solving the equation using the quadratic formula.

Coefficients

Coefficients are the numbers in front of the variables in an equation. They play a crucial role in determining the shape and position of the equation's graph. In the quadratic equation \( ax^2 + bx + c = 0 \), you will find

• \( a \), the coefficient of the quadratic term \( x^2 \)
• \( b \), the coefficient of the linear term \( x \)
• \( c \), the constant term.

For the equation \( x^2 + 6x + 7 = 0 \):

• The coefficient \( a \) is 1, meaning one \( x^2 \) term.
• The linear coefficient \( b \) is 6, which influences the slope or steepness.
• The constant \( c \) is 7, which affects where the graph of the equation intersects the y-axis.

These coefficients are incorporated into the quadratic formula to find the solutions to the equation.

Square Root

The square root is a mathematical function that, when applied, tells us what number multiplied by itself equals the given value. In the context of quadratic equations, square roots are used to simplify terms within the quadratic formula. The formula is:
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)When solving our specific equation \( x^2 + 6x + 7 = 0 \), we calculate the expression inside the square root, called the discriminant:

• \( b^2 - 4ac \)

For our coefficients \( a = 1 \), \( b = 6 \), and \( c = 7 \), the calculation becomes:

• \( 6^2 - 4 \cdot 1 \cdot 7 = 36 - 28 = 8 \)

Therefore, the square root part of the formula is \( \sqrt{8} \).
This step is crucial as it decides the nature of the roots (two distinct, one repeated, or complex). Here, as \( 8 \) is positive, we have two distinct real solutions.

Solutions to Equations

Solutions to equations are the values of the variables that make the equation true. For a quadratic equation, solutions can be found using the quadratic formula:
\( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \)For the equation \( x^2 + 6x + 7 = 0 \), substituting \( a = 1 \), \( b = 6 \), and \( c = 7 \) into the formula results in:

• \( x = \frac{-6 \pm \sqrt{8}}{2} \)

This provides two potential values:

• \( x_1 = \frac{-6 + \sqrt{8}}{2} \)
• \( x_2 = \frac{-6 - \sqrt{8}}{2} \)

These solutions tell us where the graph of the equation crosses the x-axis. They help us understand the behavior of the quadratic function, as well as its roots or zeroes.
By calculating these solutions, students can solve various problems involving motion, geometry, and even economics, using the principles of quadratic functions.