Directions: Solve the quadratic and answer the following questions. A : $(x-2)^2 = (-3x^2) + 2^2 + 25x - P$ B: $\left(10y^2 - 3^2y + \frac{2}{3}\right) \times 3 + 10y = 0$ One root of equation A is $5$. Find the product of smallest root of equation A and smallest two-digit prime number.

### Concept & Quadratic Roots To solve this, we need to find all the roots of Equation A and identify the smallest one, then multiply it by the smallest two-digit prime number. $$ \text{Product of roots} = \frac{c}{a} $$ ### Step-by-Step Solution 1. **Find Equation A and its roots:** From expanding and rearranging Equation A, we previously found: $4x^2 - 29x + P = 0$ Substituting $x=5$, we found $P=45$. The complete quadratic equation is: $4x^2 - 29x + 45 = 0$ 2. **Solve for the second root:** We can factorize the quadratic: $4x^2 - 20x - 9x + 45 = 0$ $4x(x - 5) - 9(x - 5) = 0$ $(4x - 9)(x - 5) = 0$ The roots are $x = 5$ and $x = \frac{9}{4}$. The smallest root of Equation A is $\frac{9}{4}$. 3. **Identify the smallest two-digit prime number:** The prime numbers starting from single digits are $2, 3, 5, 7, 11, 13 \dots$ The smallest two-digit prime number is $11$. 4. **Calculate the final product:** Product = $\frac{9}{4} \times 11 = \frac{99}{4}$ Convert $\frac{99}{4}$ into a mixed fraction: $99 \div 4 = 24$ with a remainder of $3$. So, $\frac{99}{4} = 24\frac{3}{4}$. ### Exam Strategy & Shortcut Instead of factorizing, use the sum or product of roots. Product of roots = $\frac{c}{a} = \frac{45}{4}$. Since one root is $5$, the other is $\frac{45}{4} \div 5 = \frac{9}{4}$. This is significantly faster! ### Common Pitfall A frequent error is misidentifying the smallest two-digit prime number as $10$ (which is composite) or $13$ (missing $11$), throwing off the final calculation. ### Final Answer Therefore, the correct answer is **$24\frac{3}{4}$**.