Here's the solution to Question 50, verifying the steps provided: Question 50: Given the equation: $x^2 + px - 10 = 0$ Assume the roots are $2$ and $-5$. Find the value of $p$. For a quadratic equation in the form $ax^2 + bx + c = 0$, the following relationships hold for its roots $r_1$ and $r_2$: Sum of roots: $r_1 + r_2 = -\frac{b}{a}$ Product of roots: $r_1 \times r_2 = \frac{c}{a}$ From the given equation $x^2 + px - 10 = 0$, we have: $a = 1$ $b = p$ $c = -10$ The assumed roots are $r_1 = 2$ and $r_2 = -5$. Step 1: Check the product of the roots. The product of the given roots is: $$ r_1 \times r_2 = 2 \times (-5) = -10 $$ Using the formula for the product of roots from the equation: $$ \frac{c}{a} = \frac{-10}{1} = -10 $$ Since the calculated product matches the formula, the assumed roots are consistent with the constant term of the equation. (Note: The student's work incorrectly showed $2 + (-5)$ for the product, but the result $-10$ was correct for the product). Step 2: Calculate the sum of the roots. The sum of the given roots is: $$ r_1 + r_2 = 2 + (-5) = -3 $$ Step 3: Use the sum of roots formula to find $p$. We know that the sum of roots is also equal to $-\frac{b}{a}$. $$ r_1 + r_2 = -\frac{b}{a} $$ Substitute the calculated sum and the values of $a$ and $b$ from the equation: $$ -3 = -\frac{p}{1} $$ $$ -3 = -p $$ Multiply both sides by $-1$: $$ p = 3 $$ Therefore, the value of $p$ is $3$. The final answer is $\boxed{p = 3}$.