Question:

Step 1: Determine the sum and product of roots $a$ and $b$ for the equation $x^2 - 3x + 1 = 0$. By Vieta's formulas: The sum of the roots is $a + b = -(-3)/1 = 3$. The product of the roots is $ab = 1/1 = 1$. Step 2: Determine the sum of the new roots $(a+2)$ and $(b+2)$. The sum of the new roots is $(a + 2) + (b + 2) = a + b + 4$. Substituting the value of $a+b$ from Step 1: $$ (a + 2) + (b + 2) = 3 + 4 = 7 $$ Step 3: Determine the product of the new roots $(a+2)$ and $(b+2)$. The product of the new roots is $(a + 2)(b + 2)$. Expanding the product: $$ (a + 2)(b + 2) = ab + 2a + 2b + 4 = ab + 2(a + b) + 4 $$ Substituting the values of $ab$ and $a+b$ from Step 1: $$ (a + 2)(b + 2) = 1 + 2(3) + 4 = 1 + 6 + 4 = 11 $$ Step 4: Form the new quadratic equation and identify $p$ and $q$. A quadratic equation with roots $\alpha$ and $\beta$ is given by $x^2 - (\alpha + \beta)x + \alpha\beta = 0$. Using the sum of new roots (7) and product of new roots (11): $$ x^2 - 7x + 11 = 0 $$ The problem states that the new quadratic equation is $x^2 + px + q = 0$. Comparing $x^2 - 7x + 11 = 0$ with $x^2 + px + q = 0$: $$ p = -7 $$ $$ q = 11 $$