If the equation $\mathrm{a}(\mathrm{b}-\mathrm{c}) \mathrm{x}^2+\mathrm{b}(\mathrm{c}-\mathrm{a}) \mathrm{x}+\mathrm{c}(\mathrm{a}-\mathrm{b})=0$ has equal roots, where $\mathrm{a}+\mathrm{c}=15$ and $\mathrm{b}=\frac{36}{5}$, then $a^2+c^2$ is equal to _________

<p>To solve the given problem, we start with the quadratic equation:</p> <p>$ a(b-c) x^2 + b(c-a) x + c(a-b) = 0 $</p> <p>Given that the roots are equal (let’s assume both roots are 1), we know the sum of the roots, $\alpha + \beta$, is twice the value of one root, which leads us to:</p> <p>$ \alpha + \beta = 2 $</p> <p>Using the formula for the sum of roots for a quadratic equation, $\alpha + \beta = -\frac{b(c-a)}{a(b-c)}$, we set this equal to 2:</p> <p>$ -\frac{b(c-a)}{a(b-c)} = 2 $</p> <p>Solving for this:</p> <p>$ -bc + ab = 2ab - 2ac \\ 2ac = ab + bc \\ 2ac = b(a + c) $</p> <p>Given that $a + c = 15$ and $b = \frac{36}{5}$, substitute these into the equation:</p> <p>$ 2ac = 15b \\ 2ac = 15 \times \frac{36}{5} = 108 \\ ac = 54 $</p> <p>Now, using the equation $a + c = 15$ and $ac = 54$, find $a^2 + c^2$:</p> <p>$ a^2 + c^2 = (a + c)^2 - 2ac = 15^2 - 2 \times 54 \\ a^2 + c^2 = 225 - 108 = 117 $</p> <p>Therefore, $a^2 + c^2$ is equal to 117.</p>