OpenSAT

We can use the relationship between the roots and coefficients of a quadratic equation. For the quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-b/a$ and the product of the roots is $c/a$. In this case, the sum of the roots, $x_1 + x_2$, is $3/2$, and the product of the roots, $x_1x_2$, is $1/2$. We want to find $x_1^2 + x_2^2$. We can square the equation $x_1 + x_2 = 3/2$ to get $x_1^2 + 2x_1x_2 + x_2^2 = 9/4$. Subtracting $2x_1x_2$ from both sides gives us $x_1^2 + x_2^2 = 9/4 - 2x_1x_2$. Substituting $x_1x_2 = 1/2$, we get $x_1^2 + x_2^2 = 9/4 - 2(1/2) = 9/4 - 1 = 5/4 = 7$. Therefore, the value of $x_1^2 + x_2^2$ is 7.