If the equation a(b – c)x² + b(c – a)x + c(a – b) = 0 has equal roots, where a + c = 15 and $\mathrm{b}=\frac{36}{5}$, then a² + c² is equal to

a(b – c) x² + b (c – a) x + c(a – b) = 0
x = 1 is root $\therefore$ other root is 1
$\begin{array}{l}\mathrm{\alpha }+\mathrm{\beta }=-\frac{\mathrm{b}(\mathrm{c}-\mathrm{a})}{\mathrm{a}(\mathrm{b}-\mathrm{c})}=2\\ \Rightarrow -\mathrm{bc}+\mathrm{ab}=2\mathrm{ab}-2\mathrm{ac}\\ \Rightarrow 2\mathrm{ac}=\mathrm{ab}+\mathrm{bc}\\ \Rightarrow 2\mathrm{ac}=\mathrm{b}(\mathrm{a}+\mathrm{c})\\ \Rightarrow 2\mathrm{ac}=15\mathrm{b}\dots \left(1\right)\\ \Rightarrow 2\mathrm{ac}=15\left(\frac{36}{5}\right)=108\end{array}$
$\Rightarrow$ ac = 54
a + c = 15
a² + c² + 2ac = 225
a² + c² = 225 – 108 = 117