Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2+bx+c=0$ with roots $r_1, r_2$, we have $r_1+r_2 = -\frac{b}{a}$ and $r_1r_2 = \frac{c}{a}$.
• Sum of Squares Identity: $(a+b)^2 = a^2 + b^2 + 2ab$, which can be rearranged to $a^2 + b^2 = (a+b)^2 - 2ab$.
• Sum of Cubes Identity: $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$.
• Forming a Quadratic Equation: A quadratic equation with roots $R_1$ and $R_2$ can be written as $x^2 - (R_1+R_2)x + (R_1R_2) = 0$.

Step-by-Step Solution

1. Analyze the Given Quadratic Equation and Find Sum/Product of Roots

• WHY: Vieta's formulas provide the essential link between the coefficients of the polynomial and its roots. We need these values to calculate expressions involving higher powers of the roots.
• The given equation is $x^2 + 2\sqrt{2}x - 1 = 0$.
• Using Vieta's formulas:
  • Sum of roots: $\alpha + \beta = -\frac{2\sqrt{2}}{1} = -2\sqrt{2}$.
  • Product of roots: $\alpha \beta = \frac{-1}{1} = -1$.

2. Calculate the Sum of Squares of Roots ($\alpha^2 + \beta^2$)

• WHY: This is a crucial intermediate step. The sum of squares allows us to calculate the sum of fourth powers, which is needed later.
• We use the identity: $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$. Rearranging, we get $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.
• Calculation: $(\alpha + \beta)^2 = (-2\sqrt{2})^2 = 8$. $2\alpha\beta = 2(-1) = -2$. So, $\alpha^2 + \beta^2 = 8 - (-2) = 8 + 2 = 10$.

3. Calculate the Sum of Fourth Powers of Roots ($\alpha^4 + \beta^4$)

• WHY: This is one of the roots of the new quadratic equation.
• We use the identity: $\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2(\alpha\beta)^2$.
• Calculation: $(\alpha^2 + \beta^2)^2 = (10)^2 = 100$. $(\alpha\beta)^2 = (-1)^2 = 1$. $2(\alpha\beta)^2 = 2(1) = 2$. So, $\alpha^4 + \beta^4 = 100 - 2 = 98$.

4. Calculate the Sum of Sixth Powers of Roots ($\alpha^6 + \beta^6$)

• WHY: This value, when scaled by $\frac{1}{10}$, forms the second root of our new quadratic equation.
• We can use the sum of cubes identity: $a^3 + b^3 = (a+b)^3 - 3ab(a+b)$. Let $a = \alpha^2$ and $b = \beta^2$.
• Then $\alpha^6 + \beta^6 = (\alpha^2)^3 + (\beta^2)^3$.
• From step 2, we know $\alpha^2 + \beta^2 = 10$. Also, $\alpha^2 \beta^2 = (\alpha\beta)^2 = (-1)^2 = 1$.
• Applying the identity: $\alpha^6 + \beta^6 = (\alpha^2 + \beta^2)^3 - 3(\alpha^2\beta^2)(\alpha^2 + \beta^2) = (10)^3 - 3(1)(10) = 1000 - 30 = 970$.

5. Determine the Roots of the New Quadratic Equation

• WHY: We are given the form of the new roots and have now calculated their values.
• The first root is $R_1 = \alpha^4 + \beta^4 = 98$.
• The second root is $R_2 = \frac{1}{10}(\alpha^6 + \beta^6) = \frac{1}{10}(970) = 97$.

6. Form the New Quadratic Equation

• WHY: To construct the final quadratic equation, we need the sum and product of its roots ($R_1$ and $R_2$).
• Sum of the new roots: $S = R_1 + R_2 = 98 + 97 = 195$.
• Product of the new roots: $P = R_1 \times R_2 = 98 \times 97 = (100-2)(100-3) = 10000 - 500 + 6 = 9506$.
• The new quadratic equation is $x^2 - Sx + P = 0$.
• Substituting the calculated values: $x^2 - 195x + 9506 = 0$.

Common Mistakes & Tips

• Sign Errors: Be extremely careful with negative signs, especially when squaring and cubing.
• Identity Selection: Ensure you select and apply the correct algebraic identities. Double-check before proceeding.
• Arithmetic Precision: Pay close attention to arithmetic calculations to avoid errors that can propagate through the solution.

Summary

This problem required us to find a new quadratic equation given the roots of an initial quadratic equation. We used Vieta's formulas to find the sum and product of the roots of the initial equation. We then calculated the sum of the fourth and sixth powers of these roots using algebraic identities. These values allowed us to determine the roots of the new quadratic equation and, consequently, form the new equation.

The final answer is $x^2 - 195x + 9506 = 0$, which corresponds to option (B).

The final answer is \boxed{x^2-195 x+9506=0}.