Key Concepts and Formulas

• Quadratic Equation and Discriminant: For a quadratic equation $Ax^2 + Bx + C = 0$, the discriminant is $D = B^2 - 4AC$. Equal roots occur when $D = 0$.
• Sum and Product of Roots: For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of roots is $-\frac{B}{A}$ and the product of roots is $\frac{C}{A}$.
• Algebraic Identity: $(a+c)^2 = a^2 + c^2 + 2ac$, which can be rearranged to $a^2 + c^2 = (a+c)^2 - 2ac$.

Step-by-Step Solution

Step 1: Identify coefficients and the condition for equal roots

The given quadratic equation is $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$. We identify the coefficients as $A = a(b-c)$, $B = b(c-a)$, and $C = c(a-b)$. Since the equation has equal roots, the discriminant must be zero: $B^2 - 4AC = 0$.

Step 2: Show that x=1 is a root

Let $P(x) = a(b-c)x^2 + b(c-a)x + c(a-b)$. Substitute $x=1$ into $P(x)$ to check if it's a root. $P(1) = a(b-c)(1)^2 + b(c-a)(1) + c(a-b) = a(b-c) + b(c-a) + c(a-b) = ab - ac + bc - ab + ac - bc = 0$ Since $P(1) = 0$, $x=1$ is a root.

Step 3: Deduce that both roots are 1

Since the problem states that the equation has equal roots, and we have found that $x=1$ is a root, both roots must be equal to 1. Let $\alpha = 1$ and $\beta = 1$.

Step 4: Apply the sum of roots formula

The sum of the roots is $\alpha + \beta = 1 + 1 = 2$. Using the formula for the sum of roots, we have $2 = -\frac{B}{A} = -\frac{b(c-a)}{a(b-c)}$

Step 5: Simplify the equation from the sum of roots

Multiply both sides by $a(b-c)$ to get: $2a(b-c) = -b(c-a)$ $2ab - 2ac = -bc + ab$ $ab + bc = 2ac$ $b(a+c) = 2ac$

Step 6: Substitute the given values of a+c and b

We are given $a+c = 15$ and $b = \frac{36}{5}$. Substitute these values into the equation $b(a+c) = 2ac$: $\frac{36}{5}(15) = 2ac$ $36(3) = 2ac$ $108 = 2ac$ $ac = 54$

Step 7: Use the algebraic identity to find a^2 + c^2

We want to find $a^2 + c^2$. We know that $(a+c)^2 = a^2 + c^2 + 2ac$. Rearranging for $a^2 + c^2$, we have: $a^2 + c^2 = (a+c)^2 - 2ac$ Substitute the values $a+c = 15$ and $ac = 54$: $a^2 + c^2 = (15)^2 - 2(54)$ $a^2 + c^2 = 225 - 108$ $a^2 + c^2 = 117$

Common Mistakes & Tips

• Testing x=1 First: Always check if $x=1$ (or $x=-1$) is a root, especially when coefficients involve differences or cyclic symmetry.
• Algebraic Manipulation: Be careful with signs and distribution when simplifying equations. A small error can lead to a wrong answer.
• Using Discriminant Directly: Using $B^2-4AC=0$ directly would be algebraically intensive and prone to error. The key is to use $x=1$ as a root.

Summary

We found that $x=1$ is a root of the quadratic equation. Since the equation has equal roots, both roots are 1. Using the sum of roots formula, we derived the relationship $b(a+c) = 2ac$. Substituting the given values for $a+c$ and $b$, we found $ac = 54$. Finally, we used the identity $a^2 + c^2 = (a+c)^2 - 2ac$ to calculate $a^2 + c^2 = 117$.

The final answer is \boxed{117}.