Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ x^{2}+6 x+1=0 $$

Short answer

The solutions are \(x = -3 + 2\sqrt{2}\) and \(x = -3 - 2\sqrt{2}\).

Step-by-step solution

Identify coefficients

The first step is to identify the coefficients in the quadratic equation. In the equation \(x^{2} + 6x + 1 = 0\), the coefficients are \(a = 1\), \(b = 6\), and \(c = 1\).

Write down the quadratic formula

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

Substitute the coefficients into the quadratic formula

Substitute \(a = 1\), \(b = 6\), and \(c = 1\) into the quadratic formula: \(x = \frac{{-6 \pm \sqrt{{6^{2} - 4 \cdot 1 \cdot 1}}}}{2 \cdot 1}\).

Simplify inside the square root

Calculate the discriminant \(b^{2} - 4ac\): \(6^{2} - 4 \cdot 1 \cdot 1 = 36 - 4 = 32\). So, the expression becomes \(x = \frac{{-6 \pm \sqrt{32}}}{2}\).

Simplify the square root

Simplify \(\sqrt{32}\): \(\sqrt{32} = 4\sqrt{2}\). Thus, the equation becomes \(x = \frac{{-6 \pm 4\sqrt{2}}}{2}\).

Simplify the final expression

Separate the expression into two fractions: \(x = \frac{-6}{2} \pm \frac{4\sqrt{2}}{2}\). Simplify each term: \(x = -3 \pm 2\sqrt{2}\).

Write the final solutions

The real solutions to the equation \(x^{2} + 6x + 1 = 0\) are \(x = -3 + 2\sqrt{2}\) and \(x = -3 - 2\sqrt{2}\).

Key concepts

quadratic equations

Quadratic equations are polynomial equations of degree 2. They have the general form:
\[ax^2 + bx + c = 0\]
Here, \(a\), \(b\), and \(c\) are constants, and \(a eq 0\) because if \(a = 0\), the equation becomes linear rather than quadratic.

Quadratic equations can be solved using various methods, such as:

• Factoring
• Completing the square
• Using the quadratic formula

The quadratic formula is a robust tool for solving any quadratic equation, but other methods like factoring require the equation to be in a specific form.

In this exercise, we are using the quadratic formula to solve \(x^2 + 6x + 1 = 0\). Let's delve deeper into understanding this formula and the key components involved in solving it.

discriminant

In the context of the quadratic formula, the discriminant is a crucial element that determines the nature of the roots (solutions) of the quadratic equation. The discriminant is the part of the quadratic formula under the square root:
\[D = b^2 - 4ac\]
Depending on the value of D, we can infer the type of solutions the quadratic equation has:

• If \(D > 0\), there are two distinct real solutions.
• If \(D = 0\), there is exactly one real solution (a repeated root).
• If \(D < 0\), there are no real solutions; instead, there are two complex solutions.

For the equation \(x^2 + 6x + 1 = 0\), we calculated the discriminant as:
\[D = 6^2 - 4 \times 1 \times 1 = 36 - 4 = 32\]
Since the discriminant is positive (\(D = 32\)), this quadratic equation has two distinct real solutions.

solving quadratic equations

Solving quadratic equations using the quadratic formula involves using a step-by-step approach. Here’s a detailed breakdown:

Step 1: Identify Coefficients
For \(x^2 + 6x + 1 = 0\), the coefficients are \(a = 1\), \(b = 6\), and \(c = 1\).

Step 2: Write Down the Quadratic Formula
The quadratic formula is:
\[x = \frac{{-b \pm \sqrt{{b^{2} - 4ac}}}}{2a}\]

Step 3: Substitute Coefficients
Plug these values into the formula:
\[x = \frac{{-6 \pm \sqrt{{6^{2} - 4 \times 1 \times 1}}}}{2 \times 1}\]

Step 4: Simplify Inside the Square Root
Calculate the discriminant:
\[6^{2} - 4 \times 1 \times 1 = 36 - 4 = 32\]

Rewrite the formula:
\[x = \frac{{-6 \pm \sqrt{32}}}{2}\]

Step 5: Simplify the Square Root
Simplify \(\sqrt{32}\):
\[\sqrt{32} = 4\sqrt{2}\]

Update the equation:
\[x = \frac{{-6 \pm 4\sqrt{2}}}{2}\]

Step 6: Simplify the Final Expression
Separate into two parts:
\[x = \frac{-6}{2} \pm \frac{4\sqrt{2}}{2}\]

Further simplification yields:
\[x = -3 \pm 2\sqrt{2}\]

Step 7: Write the Final Solutions
Therefore, the real solutions to the quadratic equation \(x^2 + 6x + 1 = 0\) are:
\[x = -3 + 2\sqrt{2}\]
and
\[x = -3 - 2\sqrt{2}\]