Problem 35 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$3 x 2+6 x+2=0$$

Step-by-Step Solution

Answer

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

Step 1: Identify the coefficients

The quadratic equation given is in the form \(ax^2 + bx + c = 0\). Here, \(a = 3\), \(b = 6\), and \(c = 2\).

Step 2: 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 values of \(x\) that satisfy the equation.

Step 3: Calculate the discriminant

The discriminant \(\Delta\) is \(b^2 - 4ac\). Plugging in the values gives \(6^2 - 4 \times 3 \times 2 = 36 - 24 = 12\). Since the discriminant is positive, we will have two real solutions.

Step 4: Apply the quadratic formula

Substitute \(a = 3\), \(b = 6\), and \(\Delta = 12\) into the quadratic formula. This gives the expression for the potential solutions: \(x = \frac{-6 \pm \sqrt{12}}{2 \times 3}\).

Step 5: Simplify the expression

Start by simplifying \(\sqrt{12}\) which equals \(2\sqrt{3}\). Thus, the solutions become \(x = \frac{-6 \pm 2\sqrt{3}}{6}\).

Step 6: Simplify the fractions

Both terms in the numerator can be divided by 2, leading to \(x = \frac{-3 \pm \sqrt{3}}{3}\). Hence, the two solutions are \(x = \frac{-3 + \sqrt{3}}{3}\) and \(x = \frac{-3 - \sqrt{3}}{3}\).

Key Concepts

Quadratic Formula

The quadratic formula is a mathematical tool used to find solutions for quadratic equations. A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \). The formula itself is given by:

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

This formula allows us to solve for \(x\) without needing to factor the quadratic equation manually.

Let's break down what each part of the formula represents:

• \(a\), \(b\), and \(c\) are the coefficients from the quadratic equation.
• \(-b\) represents the opposite of the coefficient \(b\).
• The \(\pm\) sign indicates that there are generally two solutions: one with plus, one with minus.
• \(\sqrt{b^2 - 4ac}\) is the discriminant of the equation (a key part we'll discuss more).
• \(2a\) is twice the first coefficient, forming the denominator in the formula.

Using this formula, one can find whether a quadratic equation has two solutions, one solution, or no real solutions, just by calculating.

Discriminant

The discriminant is a crucial component of the quadratic formula that determines the nature of the solutions. It is found under the square root in the quadratic formula, \( b^2 - 4ac \). The value of the discriminant reveals whether the quadratic equation's solutions are:

• Real or imaginary
• Two distinct solutions, one solution, or none

Here's how to interpret the value of the discriminant:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution (also known as a repeated root).
• If \( b^2 - 4ac < 0 \), there are no real solutions, but two imaginary ones.

In our example, \( b^2 - 4ac = 12 \), which is positive, confirming the presence of two distinct real solutions.

Real Solutions

When solving quadratic equations, one important aspect is determining the nature of the solutions. Real solutions for a quadratic equation occur when the discriminant is non-negative.

In simple terms, solutions are real if:

• The discriminant \( b^2 - 4ac \) is greater than or equal to zero.

In our specific example, after calculating the discriminant, we found it to be positive (\(12\)), which indicates that there are real solutions. Indeed, these solutions are the points where the parabola defined by the quadratic equation intersects the x-axis.

It's important to note:

• Two real solutions occur when the discriminant is positive, and the solutions represent two distinct points on the x-axis.
• One real solution means the vertex of the parabola touches the x-axis exactly once.
• No real solutions indicate that the parabola does not intersect the x-axis at all.

In our exercise's case, we calculated two distinct real solutions, ensuring the quadratic equation represented a parabola that intersects the x-axis twice.