Problem 32 - Elementary Algebra

Question

Solve. $$ 3 x 2+2 x-1=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{1}{3}\) and \(x = -1\).

Step 1: Identify the Equation Type

The given equation \(3x^2 + 2x - 1 = 0\) is a quadratic equation. It is in the form of \(ax^2 + bx + c = 0\), where \(a=3\), \(b=2\), and \(c=-1\).

Step 2: Apply the Quadratic Formula

Since this is a quadratic equation, we use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the solutions. Here, \(a=3\), \(b=2\), and \(c=-1\).

Step 3: Calculate the Discriminant

First, compute the discriminant \(b^2 - 4ac\). Substitute the values: \(b^2 - 4ac = (2)^2 - 4(3)(-1) = 4 + 12 = 16\). Since the discriminant is positive, there are two distinct real solutions.

Step 4: Solve for x

Substitute the discriminant and coefficients into the quadratic formula: \[ x = \frac{-2 \pm \sqrt{16}}{2 \times 3} = \frac{-2 \pm 4}{6} \]Calculate both solutions: 1. For \(+\), \(x = \frac{2}{6} = \frac{1}{3}\). 2. For \(-\), \(x = \frac{-6}{6} = -1\).

Step 5: Check the Solutions

Verify the solutions by substituting them back into the original equation.For \(x = \frac{1}{3}\), \[3\left(\frac{1}{3}\right)^2 + 2\left(\frac{1}{3}\right) - 1 = 0\].Simplify: \[\frac{3}{9} + \frac{2}{3} - 1 = 0\] which holds true.For \(x = -1\), \[3(-1)^2 + 2(-1) - 1 = 0\].Simplify: \[3 - 2 - 1 = 0\] which also holds true.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula allows you to directly calculate the solutions to the equation without needing to factorize. The formula itself is expressed as:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Each variable in the equation must be identified and substituted appropriately:

• \(a\) is the coefficient of \(x^2\)
• \(b\) is the coefficient of \(x\)
• \(c\) is the constant term.

Let's break down its usage by considering the example from the problem. Here \(a = 3\), \(b = 2\), and \(c = -1\). Once you've identified these values, plug them into the formula to start solving for \(x\). The beauty of the quadratic formula is that it tells you whether and how many real solutions exist by way of its discriminant, expressed as \(b^2 - 4ac\). With a bit of computation, you'll be able to solve any quadratic equation that comes your way.

Discriminant Calculation

The discriminant is a crucial part of the quadratic formula, as it determines the nature of the roots of the quadratic equation. The expression for the discriminant is \( b^2 - 4ac \). Calculating the discriminant helps us understand more about the equation's solutions before we even solve for them:

• When the discriminant is positive, the equation has two distinct real solutions.
• When it is zero, there is exactly one real solution, also called a repeated root.
• If it is negative, the quadratic equation does not have real solutions, but rather two complex solutions.

In the provided example, the calculation of the discriminant goes as follows:
\((2)^2 - 4(3)(-1) = 4 + 12 = 16\)
This positive result confirms the presence of two distinct real solutions. Recognizing the role of the discriminant can greatly simplify the process of predicting the character of solutions, saving time especially when dealing with more complicated quadratics.

Real Solutions

Real solutions to quadratic equations are numerical values that satisfy the equation when substituted back into it. To find them using the quadratic formula, we compute both possibilities from the \(\pm\) sign. Given the formula\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \], substitute the discriminant and the respective coefficients to find specific values for \(x\).
Let's go through the example:

• First, substitute the positive root: \( x = \frac{-2 + 4}{6} = \frac{2}{6} = \frac{1}{3} \).
• Then, use the negative root: \( x = \frac{-2 - 4}{6} = \frac{-6}{6} = -1 \).

These results give us two real solutions: \(x = \frac{1}{3}\) and \(x = -1\). Finally, it is beneficial to replace these numbers back into the original equation to verify their correctness.
Such an approach not only checks your work but reinforces understanding of the specific solutions found in an equation, ensuring the process and results are handled accurately.