Solve the equation 3, x, squared, plus, 9, x, plus, 1, equals, minus, 9, x3x 2 +9x+1=−9x to the nearest tenth.

1. Break Down the Problem

We need to solve the equation:

$3x^2 + 9x + 1 = -9x$

To do this, we will first rearrange the equation to have all terms on one side of the equation.

2. Relevant Concepts

• Combine like terms.
• Rearrange the equation to form a standard quadratic equation $ax^2 + bx + c = 0$.

3. Analysis and Detail

1. Rearranging the equation:

$3x^2 + 9x + 1 + 9x = 0$

This simplifies to:

$3x^2 + 18x + 1 = 0$

2. Next, we apply the quadratic formula:

$x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}$

Where $a = 3$, $b = 18$, and $c = 1$.

4. Verify and Summarize

1. Calculate the discriminant:

$b^2 - 4ac = 18^2 - 4(3)(1) = 324 - 12 = 312$

2. Now, substitute $a$, $b$, and the discriminant into the quadratic formula:

$x = \frac{{-18 \pm \sqrt{312}}}{{2 \cdot 3}} = \frac{{-18 \pm \sqrt{312}}}{6}$

3. Calculate $\sqrt{312}$:

$\sqrt{312} \approx 17.7$

4. Now we have:

$x = \frac{{-18 \pm 17.7}}{6}$

Calculating the two possible values for $x$:

• For the positive root:

$x_1 = \frac{{-18 + 17.7}}{6} \approx \frac{{-0.3}}{6} \approx -0.05 \approx -0.1 \text{ (to the nearest tenth)}$

• For the negative root:

$x_2 = \frac{{-18 - 17.7}}{6} \approx \frac{{-35.7}}{6} \approx -5.95 \approx -6.0 \text{ (to the nearest tenth)}$

Final Answer

The solutions to the equation, to the nearest tenth, are:

$x \approx -0.1 \quad \text{and} \quad x \approx -6.0$