Problem 26 - Elementary Algebra

Question

\(z^{2}+6 z+8=0\)

Step-by-Step Solution

Answer

\( z = -2 \) and \( z = -4 \)

Step 1 - Identify the quadratic equation

The given equation is a quadratic equation of the form \( az^{2} + bz + c = 0 \). For this problem, identify \( a = 1 \), \( b = 6 \), and \( c = 8 \).

Step 2 - Set up the factoring

To solve the quadratic equation by factoring, rewrite the quadratic in its factored form \( (z + p)(z + q) = 0 \). Look for two numbers that multiply to \( c = 8 \) and add to \( b = 6 \).

Step 3 - Find the factors

The two numbers that multiply to 8 and add to 6 are 2 and 4. Therefore, the factored form of the quadratic equation is \( (z + 2)(z + 4) = 0 \).

Step 4 - Solve for the roots

Set each factor equal to zero: \( z + 2 = 0 \) and \( z + 4 = 0 \). Solving these equations, we get \( z = -2 \) and \( z = -4 \).

Step 5 - Verify the solutions

Substitute \( z = -2 \) and \( z = -4 \) back into the original equation to verify the solutions. Both should satisfy the equation \( z^{2} + 6z + 8 = 0 \).

Key Concepts

factoring quadratics

When solving quadratic equations, one effective method is factoring. Factoring means expressing the quadratic equation in a product of simpler binomials. For example, take the equation from our exercise: \( z^2 + 6z + 8 = 0 \).

First, note that this equation follows the standard quadratic form \( az^2 + bz + c = 0 \), where in our case, \( a = 1 \), \( b = 6 \), and \( c = 8 \).

Think of two numbers that multiply to \( c \) (which is 8) and add up to \( b \) (which is 6). Here, the numbers 2 and 4 fit perfectly because:

• 2 * 4 = 8
• 2 + 4 = 6

So, we can rewrite the quadratic equation in its factored form as:

\( (z + 2)(z + 4) = 0 \)

This factored form is critical because it allows us to solve for the variable \( z \) by setting each binomial equal to zero.

roots of quadratic equations

The solutions to a quadratic equation, also known as the roots, can be found once the quadratic is factored. In our exercise, our factored equation was:

\( (z + 2)(z + 4) = 0 \)

To find the roots, set each factor equal to zero and solve for \( z \):

• \( z + 2 = 0 \) → \( z = -2 \)
• \( z + 4 = 0 \) → \( z = -4 \)

So, the roots of the quadratic equation \( z^2 + 6z + 8 \) are \( z = -2 \) and \( z = -4 \). These values of \( z \) are the points where the quadratic graph intersects the x-axis.

verifying solutions of quadratics

After finding the potential solutions of a quadratic equation, it's crucial to verify them. For the equation \( z^2 + 6z + 8 = 0 \), the solutions were \( z = -2 \) and \( z = -4 \).
To verify, substitute these values back into the original equation and check if they satisfy the equation. Let's substitute \( z = -2 \):
\( (-2)^2 + 6(-2) + 8 = 0 \) becomes:

4 - 12 + 8 = 0

0 = 0

Which is true. Next, substitute \( z = -4 \): \( (-4)^2 + 6(-4) + 8 = 0 \) becomes:

16 - 24 + 8 = 0

0 = 0Which is also true.

Because both substitutions both produce true statements, \( z = -2 \) and \( z = -4 \) are indeed the correct solutions to the quadratic equation. Verifying solutions ensures that your factoring and calculations are correct.

quadratic formula

Sometimes, factoring quadratics is not straightforward. In such cases, the quadratic formula is an incredibly useful tool. The quadratic formula is:

© = \(ax^2 + bx + c = 0 \), the roots can be found using:

\( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

Here is how you use the quadratic formula for the same equation:

For \( z^2 + 6z + 8 = 0 \), with \( a = 1 \), \( b = 6 \), and \( c = 8 \), substitute these values into the quadratic formula:


\( z = \frac{-6 \pm \sqrt{6^2 - 4*1*8}}{2*1} \), which simplifies to:

\( z = \frac{-6 \pm \sqrt{36 - 32}}{2} \ =~ \frac{-6 \pm 2}{2}\).

Thus, we get the roots:





\( z = \frac{-6 + 2}{2} = -2 \)

\( z = \frac{-6 - 2}{2} = -4 \)The quadratic calculator formula can be an important solution method, especially when factoring is complex or too many factors exist.