1. **Problem statement:** Given that one root of the quadratic equation $x^2 - 3x + a = 0$ is $x=6$, find the value of $a$ and the other root. 2. **Formula and rules:** For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$. 3. **Step 1: Use the given root to find $a$.** Substitute $x=6$ into the equation: $$6^2 - 3 \times 6 + a = 0$$ $$36 - 18 + a = 0$$ $$18 + a = 0$$ $$a = -18$$ 4. **Step 2: Find the other root.** The sum of roots is: $$x_1 + x_2 = -\frac{b}{a} = -\frac{-3}{1} = 3$$ Given $x_1 = 6$, then: $$6 + x_2 = 3$$ $$x_2 = 3 - 6 = -3$$ 5. **Final answer:** $$a = -18$$ $$\text{Other root} = -3$$