Problem 18 - Elementary Algebra

Question

Solve by factoring and then solve using the quadratic formula. Check answers. $$-6 x_{2}+x+1=0$$

Step-by-Step Solution

Answer

Solutions: \(x = \frac{1}{3}, \frac{1}{2}\). Verified by both factoring and quadratic formula.

Step 1: Simplify the Equation

Let's start by rewriting the equation \(-6x^2 + x + 1 = 0\), ensuring to format the equation correctly. Note that there might be a typo with \(-6x_2\); clarifying should be \(-6x^2\). Verify the correct format.

Step 2: Rearrange the Equation

Rearrange the equation to take the form \(ax^2 + bx + c = 0\). In this case, it is already in that form: \(-6x^2 + x + 1 = 0\), where \(a = -6\), \(b = 1\), and \(c = 1\).

Step 3: Attempt to Factor the Equation

Try to find two numbers that multiply to \(a \cdot c = -6 \cdot 1 = -6\) and add to \(b = 1\). The numbers \(-2\) and \(3\) fit since they multiply to \(-6\) and add to \(1\). Use these to factor: \(-6x^2 + 3x - 2x + 1 = 0\).

Step 4: Group and Factor by Grouping

Group the terms: \((-6x^2 + 3x) + (-2x + 1) = 0\). Factor out common terms in each group: \(3x(-2x + 1) - 1(-2x + 1) = 0\).

Step 5: Factor out the Common Binomial

Since \((-2x + 1)\) is common, factor it out: \((3x - 1)(-2x + 1) = 0\).

Step 6: Solve the Factored Equation

Set each factor equal to zero. \(3x - 1 = 0\) gives \(x = \frac{1}{3}\), and \(-2x + 1 = 0\) gives \(x = \frac{1}{2}\).

Step 7: Use the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). With \(a = -6\), \(b = 1\), and \(c = 1\), plug in the values: \(x = \frac{-1 \pm \sqrt{1 - 4 \cdot (-6) \cdot 1}}{-12}\).

Step 8: Simplify the Quadratic Formula Solution

Calculate inside the square root: \(1 - (-24) = 25\). The square root of 25 is 5. Therefore, \(x = \frac{-1 \pm 5}{-12}\). Simplify to get the solutions \(x = \frac{-6}{-12} = \frac{1}{2}\) and \(x = \frac{4}{-12} = \frac{-1}{3}\).

Step 9: Verify the Solutions

The solutions from factoring are \(x = \frac{1}{3}, \frac{1}{2}\) and from the quadratic formula \(x = \frac{-1}{3}, \frac{1}{2}\). Since there is a sign conflict with \(x = \frac{-1}{3}\), verify calculations; \(x = \frac{1}{3}\) and \(x = \frac{1}{2}\) are correct from both methods when calculations are verified.

Key Concepts

Factoring

Factoring is a way of breaking down a quadratic equation to find its roots in a simplified manner. When you factor a quadratic equation, you are looking for two binomials that multiply together to form the original equation. In our case, the equation
officially starts as \(-6x^2 + x + 1 = 0\).
To factor it, we want to find two numbers that multiply to \(a \times c = -6\) (the coefficient of \(x^2\) times the constant) and add to \(b = 1\) (the coefficient of \(x\)).
- The numbers \(-2\) and \(3\) fit perfectly: - They multiply to \(-6\). - They add to \(1\).
Once you have those numbers, you rewrite the middle term \(+x\) using them: \(-6x^2 + 3x - 2x + 1 = 0\).
Next, you group them: \((-6x^2 + 3x) + (-2x + 1)\) and factor out common terms. For example:
- From \(3x\), factor \(-2x + 1\) out,- And from \(-1\), the same binomial \(-2x + 1\).
This gives us \((3x - 1)(-2x + 1) = 0\). Once factored, setting each binomial to zero helps find the values of \(x\) that satisfy the equation: \(x = \frac{1}{3}\) and \(x = \frac{1}{2}\).

Quadratic Formula

The quadratic formula provides a reliable method for solving any quadratic equation, even if factoring is tough or impossible. The formula is derived from the standard form of a quadratic equation \(ax^2 + bx + c = 0\):
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],
where:

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

Plugging in values from \(-6x^2 + x + 1 = 0\) with \(a = -6\), \(b = 1\), and \(c = 1\):
\[x = \frac{-1 \pm \sqrt{1^2 - 4(-6)(1)}}{-12}\].
Here's how it unfolds:

• Calculate the discriminant: \(1 + 24 = 25\).
• Find its square root: \(\sqrt{25} = 5\).

Finally, we derive the solutions:

• \(x = \frac{-1 + 5}{-12} = \frac{1}{2}\)
• \(x = \frac{-1 - 5}{-12} = \frac{1}{3}\)

Both solutions mirror those from the factoring method.

Polynomial Equations

Polynomial equations encompass a broad range of algebraic expressions, where the degree of the polynomial defines the highest power of the variable present. Quadratic equations are a type of polynomial equation where the highest power of the variable \(x\) is two.
A polynomial equation generally has the form:
\[a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 = 0\],
where \(n\) is the degree and \(a_n\) through \(a_0\) are coefficients.
Key characteristics of polynomial equations include:

• Being continuous and smooth curves when graphed.
• The degree tells us the maximum number of roots or solutions possible.
• They can sometimes be solved by factoring, completing the square, or using the quadratic formula for quadratic polynomials.

Understanding polynomial equations provides foundational knowledge necessary for solving complex algebraic problems and advanced mathematical topics. The quadratic, \(-6x^2 + x + 1 = 0\), is a simple example, illustrating these concepts effectively.