Problem 1 - Algebra 2 and Trigonometry

Question

Noah said that the solutions of Example \(1, \frac{2+\sqrt{6}}{2}\) and \(\frac{2-\sqrt{6}}{2},\) could have been written as \(\frac{2+\sqrt{6}}{2}=1+\sqrt{6}\) and \(\frac{-}{2}-\sqrt{6}=1-\sqrt{6} .\) Do you agree with Noah? Explain why or why not.

Step-by-Step Solution

Answer

No, Noah's expressions are incorrect; proper division wasn't applied.

Step 1: Understand the Question

Noah proposes that the expression for the solution \( \frac{2+\sqrt{6}}{2} \) can be written as \( 1+\sqrt{6} \) and similarly for the other term. We need to verify if these transformations are mathematically valid.

Step 2: Simplify the Expressions

First, let's simplify \( \frac{2+\sqrt{6}}{2} \). Divide each term in the numerator by 2: \( \frac{2}{2} + \frac{\sqrt{6}}{2} = 1 + \frac{\sqrt{6}}{2} \). This does not equal \( 1 + \sqrt{6} \). Thus, Noah's statement for \( \frac{2+\sqrt{6}}{2} \) is incorrect.

Step 3: Check the Second Expression

Similarly, for \( \frac{2-\sqrt{6}}{2} \), we divide each term by 2: \( \frac{2}{2} - \frac{\sqrt{6}}{2} = 1 - \frac{\sqrt{6}}{2} \). This is not equal to \( 1 - \sqrt{6} \). Hence, Noah's re-expression of the second solution as \( 1-\sqrt{6} \) is also incorrect.

Step 4: Conclusion

Both expressions Noah suggested are incorrect because he did not properly divide each term in the numerators by 2. The correct simplifications were shown in the previous steps.

Key Concepts

Quadratic Solutions

Quadratic solutions are the values of the variable that satisfy a quadratic equation. A quadratic equation is typically in the form \( ax^2 + bx + c = 0 \). Solving it involves finding the values of \( x \) that make the equation true. Common methods include factoring, completing the square, and the quadratic formula. The most popular approach is the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]This formula calculates the roots by using the coefficients \( a \), \( b \), and \( c \) from the quadratic equation. The discriminant \( b^2 - 4ac \) determines the nature of the roots:

• If it's positive, there are two distinct real roots.
• If it's zero, there is exactly one real root.
• If it's negative, there are two complex roots.

Understanding quadratic solutions is crucial for solving these equations efficiently. In our example, when solutions are expressed as fractions with radicals, it's important not to overlook the proper distribution of factors across both terms of the fraction.

Radical Expressions

Radical expressions involve roots, such as square roots, cube roots, etc. They appear in equations when solving for unknowns and sometimes need to be simplified for clearer understanding. Simplifying a radical expression involves factoring out powers of numbers that are perfect powers of the root. For example, simplifying the square root of 18 involves:

• Breaking it into factors of 9 and 2: \( \sqrt{18} = \sqrt{9 \times 2} \)
• Simplifying to \( 3\sqrt{2} \) by taking \( \sqrt{9} \) out as 3.

More complex expressions, like \( \frac{2+\sqrt{6}}{2} \), involve separating terms in the numerator and simplifying each individually. It's crucial to handle each part accurately to avoid incorrect conclusions, as seen in our original problem example.

Simplifying Fractions

Simplifying fractions is reducing them to their simplest form by dividing the numerator and the denominator by their greatest common factor. This ensures that both numbers share no common divisors other than one. When simplifying fractions involving radicals, each component of the fraction must be divided separately. Take \( \frac{2+\sqrt{6}}{2} \):

• Split it into two distinct terms: \( \frac{2}{2} + \frac{\sqrt{6}}{2} \)
• Simplify to \( 1 + \frac{\sqrt{6}}{2} \)

This method ensures clarity and avoids errors. Proper simplification of fractions with radicals is essential to derive the correct solution and is a key aspect of algebraic simplification.