If $a(b-c)x^2 + b(c-a)x + c(a-b) = 0$ has equal roots and $a+c=15$, $b = \dfrac{36}{5}$, then $a^2+c^2$ is equal to
Step-by-Step Solution
Key Concept: Observe that $x=1$ is always a root (coefficients sum to zero); equal roots forces both roots to equal 1, which via Vieta's product gives $2ac = b(a+c)$.
Sum of coefficients $= a(b-c)+b(c-a)+c(a-b) = 0$, so $x=1$ is always a root.
For equal roots, both roots equal 1. By Vieta's, product of roots $= \dfrac{c(a-b)}{a(b-c)} = 1$, giving
$$c(a-b) = a(b-c) \Rightarrow ca - cb = ab - ac \Rightarrow 2ac = b(a+c).$$
Substituting $a+c=15$ and $b=36/5$:
$$2ac = \frac{36}{5}\times15 = 108 \Rightarrow ac = 54.$$
$$a^2+c^2 = (a+c)^2 - 2ac = 225 - 108 = 117.$$
Correct Answer: 117