Step 1: Find the sum and product of the roots $\alpha$ and $\beta$ for the given equation $2x^2 + 4x + 3 = 0$. For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha\beta = \frac{c}{a}$. Given $2x^2 + 4x + 3 = 0$, we have $a=2$, $b=4$, $c=3$. Sum of roots: $$ \alpha + \beta = -\frac{4}{2} = -2 $$ Product of roots: $$ \alpha\beta = \frac{3}{2} $$ Step 2: Find the sum of the new roots, which are $\alpha - 1$ and $\beta - 1$. Let the new roots be $r_1 = \alpha - 1$ and $r_2 = \beta - 1$. Sum of new roots: $$ r_1 + r_2 = (\alpha - 1) + (\beta - 1) = \alpha + \beta - 2 $$ Substitute the value of $\alpha + \beta$: $$ r_1 + r_2 = -2 - 2 = -4 $$ Step 3: Find the product of the new roots. Product of new roots: $$ r_1r_2 = (\alpha - 1)(\beta - 1) = \alpha\beta - \alpha - \beta + 1 = \alpha\beta - (\alpha + \beta) + 1 $$ Substitute the values of $\alpha\beta$ and $\alpha + \beta$: $$ r_1r_2 = \frac{3}{2} - (-2) + 1 = \frac{3}{2} + 2 + 1 = \frac{3}{2} + 3 = \frac{3}{2} + \frac{6}{2} = \frac{9}{2} $$ Step 4: Form the new quadratic equation. A quadratic equation with roots $r_1$ and $r_2$ is given by $x^2 - (r_1 + r_2)x + r_1r_2 = 0$. Substitute the sum and product of the new roots: $$ x^2 - (-4)x + \frac{9}{2} = 0 $$ $$ x^2 + 4x + \frac{9}{2} = 0 $$ To eliminate the fraction, multiply the entire equation by 2: $$ 2(x^2 + 4x + \frac{9}{2}) = 2(0) $$ $$ 2x^2 + 8x + 9 = 0 $$ Comparing this with the given options, option (b) matches our result. The final answer is $\boxed{\text{2x^2 + 8x + 9 = 0}}$.