Key Concepts and Formulas

• For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$.
• Change of base formula for logarithms: $\log_a b = \frac{\log_c b}{\log_c a}$. In particular, $\log_a b = \frac{1}{\log_b a}$.
• Exponential identity: $a^{\log_a b} = b$ and $(a^m)^n = a^{mn}$.

Step-by-Step Solution

1. Simplify the Given Quadratic Equation

The given equation is: $x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0$ We aim to simplify the coefficients of this equation using properties of logarithms and exponents.

2. Simplify the Coefficient of x

The coefficient of $x$ is $5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}$. Let $k = \log_3 5$. Then $\log_5 3 = \frac{1}{\log_3 5} = \frac{1}{k}$. Consider the term $3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}} = 3^{\sqrt{k}} - 5^{\sqrt{1/k}}$. Since $5 = 3^{\log_3 5} = 3^k$, we can write $5^{\sqrt{1/k}} = (3^k)^{\sqrt{1/k}} = 3^{k \cdot \sqrt{1/k}} = 3^{k \cdot \frac{1}{\sqrt{k}}} = 3^{\sqrt{k}}$. Therefore, $3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}} = 3^{\sqrt{k}} - 3^{\sqrt{k}} = 0$. The coefficient of $x$ simplifies to $5 + 0 = 5$.

3. Simplify the Constant Term

The constant term is $3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)$. Let $m = \log_3 5$. Then $\log_5 3 = \frac{1}{m}$. Consider the term $3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} = 3^{m^{1/3}} - 5^{(1/m)^{2/3}}$. Substitute $5 = 3^m$, then $5^{(1/m)^{2/3}} = (3^m)^{(1/m)^{2/3}} = 3^{m \cdot (1/m)^{2/3}} = 3^{m \cdot (1/m^{2/3})} = 3^{m^{1 - 2/3}} = 3^{m^{1/3}}$. So, $3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}} = 3^{m^{1/3}} - 3^{m^{1/3}} = 0$. The expression inside the parenthesis of the constant term becomes $0 - 1 = -1$. The constant term simplifies to $3 \times (-1) = -3$.

4. Write the Simplified Quadratic Equation

The original quadratic equation simplifies to: $x^2 - 5x - 3 = 0$

5. Determine the Sum and Product of Original Roots

For the equation $x^2 - 5x - 3 = 0$, let the roots be $\alpha$ and $\beta$.

• Sum of roots: $\alpha + \beta = -(-5)/1 = 5$.
• Product of roots: $\alpha\beta = -3/1 = -3$.

6. Define the New Roots

The new roots are given as $p = \alpha + \frac{1}{\beta}$ and $q = \beta + \frac{1}{\alpha}$.

7. Calculate the Sum of the New Roots

$p+q = \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right)$ $p+q = (\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$ To find $\frac{1}{\alpha} + \frac{1}{\beta}$, we combine the fractions: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\beta + \alpha}{\alpha\beta}$ Substitute the values of $\alpha + \beta$ and $\alpha\beta$: $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{5}{-3} = -\frac{5}{3}$ Now, substitute this back into the sum of new roots: $p+q = (\alpha + \beta) + \frac{\alpha + \beta}{\alpha\beta} = 5 + \left(-\frac{5}{3}\right) = \frac{15 - 5}{3} = \frac{10}{3}$

8. Calculate the Product of the New Roots

$pq = \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right)$ Expand the product: $pq = \alpha\beta + \alpha\left(\frac{1}{\alpha}\right) + \left(\frac{1}{\beta}\right)\beta + \left(\frac{1}{\beta}\right)\left(\frac{1}{\alpha}\right)$ $pq = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta}$ $pq = \alpha\beta + 2 + \frac{1}{\alpha\beta}$ Substitute the values of $\alpha\beta$: $pq = (-3) + 2 + \frac{1}{(-3)}$ $pq = -1 - \frac{1}{3} = -\frac{4}{3}$

9. Form the New Quadratic Equation

The new quadratic equation is $x^2 - (p+q)x + pq = 0$. Substitute the calculated sum ($p+q = \frac{10}{3}$) and product ($pq = -\frac{4}{3}$): $x^2 - \left(\frac{10}{3}\right)x + \left(-\frac{4}{3}\right) = 0$ To eliminate the fractions, multiply the entire equation by 3: $3 \left(x^2 - \frac{10}{3}x - \frac{4}{3}\right) = 3 \times 0$ $3x^2 - 10x - 4 = 0$

10. Transform the Equation to Match the Correct Answer

Since the stated "Correct Answer" is (A) $3x^2 - 20x - 12 = 0$, we need to work backwards and check for an error in the original question or answer key. Let's ASSUME that the correct new roots are $p = 3\alpha + \frac{1}{\beta}$ and $q = 3\beta + \frac{1}{\alpha}$.

Then $p+q = 3(\alpha+\beta) + \frac{\alpha+\beta}{\alpha\beta} = 3(5) + \frac{5}{-3} = 15 - \frac{5}{3} = \frac{40}{3}$. And $pq = (3\alpha + \frac{1}{\beta})(3\beta + \frac{1}{\alpha}) = 9\alpha\beta + 3 + 3 + \frac{1}{\alpha\beta} = 9(-3) + 6 + \frac{1}{-3} = -27 + 6 - \frac{1}{3} = -21 - \frac{1}{3} = -\frac{64}{3}$.

The quadratic then becomes $x^2 - \frac{40}{3}x - \frac{64}{3} = 0$, or $3x^2 - 40x - 64 = 0$. This doesn't match (A).

Let's try the roots $p = \alpha + \frac{1}{3\beta}$ and $q = \beta + \frac{1}{3\alpha}$. $p+q = (\alpha+\beta) + \frac{1}{3}\frac{\alpha+\beta}{\alpha\beta} = 5 + \frac{1}{3} \frac{5}{-3} = 5 - \frac{5}{9} = \frac{40}{9}$. $pq = (\alpha+\frac{1}{3\beta})(\beta+\frac{1}{3\alpha}) = \alpha\beta + \frac{1}{3} + \frac{1}{3} + \frac{1}{9\alpha\beta} = -3 + \frac{2}{3} + \frac{1}{9(-3)} = -3 + \frac{2}{3} - \frac{1}{27} = \frac{-81+18-1}{27} = \frac{-64}{27}$. The quadratic is $x^2 - \frac{40}{9}x - \frac{64}{27} = 0$, or $27x^2 - 120x - 64 = 0$. This doesn't match (A).

Let's assume there was a typo in the original equation, and the correct equation is actually $x^2 - (5 + 0)x - \frac{4}{3} = 0$, or $3x^2 - 15x - 4 = 0$, then $\alpha + \beta = 5$ and $\alpha\beta = -\frac{4}{3}$. In this case, we want the roots $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$. The sum is $5 + \frac{5}{-4/3} = 5 - \frac{15}{4} = \frac{5}{4}$. The product is $-\frac{4}{3} + 2 + \frac{-3}{4} = \frac{-16 + 24 - 9}{12} = \frac{-1}{12}$. So the quadratic is $x^2 - \frac{5}{4}x - \frac{1}{12} = 0$, or $12x^2 - 15x - 1 = 0$.

Let's assume the roots are $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$, and the quadratic is $3x^2 - 20x - 12 = 0$. Then the sum of roots is $\frac{20}{3}$ and the product is $-4$. Then $\alpha+\beta + \frac{\alpha+\beta}{\alpha\beta} = \frac{20}{3}$ and $\alpha\beta + 2 + \frac{1}{\alpha\beta} = -4$. So $\alpha\beta = -3$ is close, but doesn't work.

Let's try the roots $3\alpha$ and $3\beta$. Then $\alpha+\beta = 5$ and $\alpha\beta = -3$, so $3\alpha + 3\beta = 15$ and $9\alpha\beta = -27$. So this is wrong.

Let's assume the constant term is $16/3$, not -12/3. Then $3x^2 - 20x + 16 = 0$. Then the roots are $3\alpha$ and $\beta/3$ is not correct.

We are told that the correct answer is (A) $3x^2 - 20x - 12 = 0$. This means $p+q = 20/3$ and $pq = -4$. If $\alpha + \beta = 5$ and $\alpha \beta = -3$, then $p+q = \alpha+\beta + \frac{\alpha+\beta}{\alpha \beta} = 5 + \frac{5}{-3} = \frac{10}{3}$. $pq = \alpha \beta + 2 + \frac{1}{\alpha \beta} = -3 + 2 - \frac{1}{3} = -\frac{4}{3}$. These are NOT the roots.

Instead of assuming $\alpha+\frac{1}{\beta}$ and $\beta+\frac{1}{\alpha}$, let's assume the roots are $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$. Then $\frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2 - 2\alpha \beta}{\alpha \beta} = \frac{25 + 6}{-3} = -\frac{31}{3}$. And $\frac{\alpha}{\beta} \cdot \frac{\beta}{\alpha} = 1$. So the equation is $x^2 + \frac{31}{3} x + 1 = 0$, or $3x^2 + 31x + 3 = 0$.

It seems there is an error in the problem statement or the correct answer. The correct derivation leads to $3x^2 - 10x - 4 = 0$.

Common Mistakes & Tips

• Be careful with signs when applying Vieta's formulas.
• Double-check the simplification of logarithmic and exponential expressions. A small error there can propagate through the entire solution.
• When transforming roots, systematically calculate the sum and product of the new roots.

Summary

The problem involves simplifying a quadratic equation with complex coefficients, finding the sum and product of its roots, and then finding a new quadratic equation whose roots are transformations of the original roots. The simplification relies on properties of logarithms and exponents. Based on the given information, the simplified quadratic equation is $x^2 - 5x - 3 = 0$, and the transformed quadratic equation is $3x^2 - 10x - 4 = 0$. However, this contradicts the provided "Correct Answer: A". There appears to be an error either in the original problem statement or the answer key, as a rigorous derivation leads to the equation $3x^2 - 10x - 4 = 0$.

Final Answer

The final answer is \boxed{3x^2 - 10x - 4 = 0}, which corresponds to option (B). However, the stated correct answer is (A).