Key Concepts and Formulas

• Discriminant: For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is $\Delta = B^2 - 4AC$. If $\Delta = 0$, the quadratic equation has one repeated real root.
• Vieta's Formulas: For a quadratic equation $Ax^2 + Bx + C = 0$ with roots $r_1$ and $r_2$, the sum of the roots is $r_1 + r_2 = -\frac{B}{A}$ and the product of the roots is $r_1 r_2 = \frac{C}{A}$.
• Algebraic Identity: $(\alpha + \beta)^2 = \alpha^2 + 2\alpha\beta + \beta^2$, which can be rearranged to $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$.

Step 1: Analyze the First Equation and Find a Relationship between a and b

The first equation $ax^2 - 2bx + 15 = 0$ has a repeated root $\alpha$. This implies the discriminant is zero.

The coefficients are $A = a$, $B = -2b$, and $C = 15$. Applying the discriminant condition: $(-2b)^2 - 4(a)(15) = 0$ $4b^2 - 60a = 0$ $b^2 = 15a \quad (*)$

Why? Setting the discriminant to zero allows us to establish a crucial relationship between a and b, connecting the two equations.

Step 2: Find the Value of the Repeated Root $\alpha$ in Terms of b

Since $\alpha$ is a repeated root, the sum of the roots is $2\alpha$. Using Vieta's formulas on the first equation: $2\alpha = -\frac{-2b}{a} = \frac{2b}{a}$ $\alpha = \frac{b}{a}$

Substitute $a = \frac{b^2}{15}$ (from equation (*)) into the expression for $\alpha$: $\alpha = \frac{b}{\frac{b^2}{15}} = \frac{15b}{b^2} = \frac{15}{b}$

Why? Expressing $\alpha$ in terms of b is vital as it links the first equation to the second, where $\alpha$ is also a root.

Step 3: Substitute $\alpha$ into the Second Equation and Solve for $b^2$

The second equation is $x^2 - 2bx + 21 = 0$, with roots $\alpha$ and $\beta$. Since $\alpha$ is a root, it must satisfy the equation: $\alpha^2 - 2b\alpha + 21 = 0$

Substitute $\alpha = \frac{15}{b}$ into the equation: $\left(\frac{15}{b}\right)^2 - 2b\left(\frac{15}{b}\right) + 21 = 0$ $\frac{225}{b^2} - 30 + 21 = 0$ $\frac{225}{b^2} - 9 = 0$ $\frac{225}{b^2} = 9$ $b^2 = \frac{225}{9} = 25$

Why? Substituting the value of $\alpha$ allows us to determine a numerical value for $b^2$, which is crucial for finding the final answer.

Step 4: Calculate $\alpha^2 + \beta^2$ using Vieta's Formulas and the Value of $b^2$

For the second equation, $x^2 - 2bx + 21 = 0$, with roots $\alpha$ and $\beta$, Vieta's formulas give: $\alpha + \beta = -\frac{-2b}{1} = 2b$ $\alpha\beta = \frac{21}{1} = 21$

Using the algebraic identity: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ $\alpha^2 + \beta^2 = (2b)^2 - 2(21)$ $\alpha^2 + \beta^2 = 4b^2 - 42$

Substitute $b^2 = 25$: $\alpha^2 + \beta^2 = 4(25) - 42$ $\alpha^2 + \beta^2 = 100 - 42$ $\alpha^2 + \beta^2 = 58$

Why? Using Vieta's formulas and the algebraic identity allows us to calculate $\alpha^2 + \beta^2$ efficiently, without needing to find the individual values of $\alpha$ and $\beta$.

Common Mistakes & Tips

• Remember to use the discriminant condition correctly for repeated roots ($B^2 - 4AC = 0$).
• Vieta's formulas are powerful tools for relating roots and coefficients, especially for symmetric expressions.
• Avoid unnecessary calculations of individual root values; instead, focus on using identities.

Summary

We used the discriminant condition on the first equation to relate a and b. Then, we expressed the repeated root $\alpha$ in terms of b and substituted it into the second equation to find $b^2$. Finally, we used Vieta's formulas on the second equation and an algebraic identity to calculate $\alpha^2 + \beta^2$.

The final answer is \boxed{58}, which corresponds to option (B).