Question

Find the real solutions, if any, of each equation. Use the quadratic formula. $$ 3 x(x+2)=1 $$

Short answer

The real solutions are \[ x = -1 + \frac{2\sqrt{3}}{3} \] and \[ x = -1 - \frac{2\sqrt{3}}{3} \].

Step-by-step solution

- Write the equation in standard quadratic form

To use the quadratic formula, first write the given equation in the form of \( ax^2 + bx + c = 0 \). Start by expanding and simplifying the given equation: \[ 3x(x+2) = 1 \] becomes \[ 3x^2 + 6x = 1 \]. Now, move all terms to one side to set the equation to zero: \[ 3x^2 + 6x - 1 = 0 \].

- Identify coefficients

With the equation in standard form \( 3x^2 + 6x - 1 = 0 \), identify the coefficients: \(a = 3\), \(b = 6\), and \(c = -1\).

- Apply the quadratic formula

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the solutions. Substitute the values of the coefficients: \( a = 3 \), \( b = 6 \), and \( c = -1 \) into the formula: \[ x = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \].

- Simplify inside the square root

Calculate the discriminant \( b^2 - 4ac \): \( 6^2 - 4 \cdot 3 \cdot (-1) = 36 + 12 = 48 \). So the quadratic formula simplifies to: \[ x = \frac{-6 \pm \sqrt{48}}{6} \].

- Simplify the expression

Since \( \sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3} \), substitute back into the equation: \[ x = \frac{-6 \pm 4\sqrt{3}}{6} = \frac{-6}{6} \pm \frac{4\sqrt{3}}{6} \]. This becomes \[ x = -1 \pm \frac{2\sqrt{3}}{3} \].

- Write the final solutions

The real solutions to the equation \( 3x(x + 2) = 1 \) are: \[ x = -1 + \frac{2\sqrt{3}}{3} \] and \[ x = -1 - \frac{2\sqrt{3}}{3} \].

Key concepts

Quadratic Formula

The quadratic formula is a powerful tool to find the roots of a quadratic equation, which is any equation of the form \( ax^2 + bx + c = 0 \). The formula is written as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \].
Here's how you use it:
First, identify the coefficients \(a\), \(b\), and \(c\) from your equation.
These coefficients are the numbers in front of the variables \(x^2\), \(x\), and the constant term, respectively.
Next, plug these values into the quadratic formula.
The terms \(b^2 - 4ac\) inside the square root are very important.
This is called the discriminant, and it tells us about the nature of the roots.
If the discriminant is positive, there are two real and distinct solutions.
If it's zero, there is exactly one real solution.
If the discriminant is negative, the solutions are complex or imaginary.
Let's go through the step-by-step process to better understand.

Discriminant Calculation

The discriminant is a part of the quadratic formula found inside the square root and is given by \( b^2 - 4ac \).
This value tells us if we will get real or complex solutions:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution.
• If \( b^2 - 4ac < 0 \), there are two complex solutions.

In our example, we calculated the discriminant as: \begin{align*} 6^2 - 4 \times 3 \times -1 &= 36 + 12 \ &= 48 \. So, \(48>0\), which means we have two distinct real solutions.
It is always good to calculate the discriminant first to understand the nature of the roots before plugging values into the quadratic formula.

Simplifying Square Roots

Once we have identified the values of \(a\), \(b\), and \(c\), and calculated the discriminant, we often need to simplify the square root part of the quadratic formula.
For example, in our solution, we ended up with \( \sqrt{48} \).
To simplify this, we find the largest square factor of 48.
We know that 48 can be written as 16 * 3, and since 16 is a perfect square (\( \sqrt{16} = 4 \)), we have:
\( \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3} \).
Now, we substitute this back into our formula: \[ x = \frac{-6 \pm 4\sqrt{3}}{6} = -1 \pm \frac{2\sqrt{3}}{3} \].
Always break down your square root into simpler parts to make calculations easy and to express your final answer in the simplest form possible.