Question 28

Explanation

This system of linear equations can be written in matrix form $AX = B$, where

Cramer's rule can be used because the determinant of the coefficient matrix $\det(A) = -4 - 3 = -7 \neq 0$ (determinant is nonzero). Cramer's rule states that for the system $AX = B$ with $\det(A) \neq 0$, the solution is

where $A_x$ is the matrix $A$ with the first column replaced by $B$, and $A_y$ is the matrix $A$ with the second column replaced by $B$.

By using Cramer's rule, we obtain

Next, substitute the values of $x$ and $y$ into $x\sqrt{y} + 3 > 0$ to obtain

To simplify $\sqrt{(a + 1)^2}$, note that $\sqrt{(a + 1)^2} = |a + 1|$. For $\sqrt{y}$ to be defined, we must have $y \geq 0$, that is $(a + 1)^2 \geq 0$, which is always true.

Choose $a < -1$ so that the form $\sqrt{(a + 1)^2} = -(a + 1)$, thus we obtain

To find the value of $a$, we can use the quadratic formula. The quadratic equation $a^2 + 2a - 2 = 0$ has roots

For the inequality $a^2 + 2a - 2 < 0$, since the coefficient of $a^2$ is positive, the inequality is satisfied for values of $a$ between its two roots, that is $-1 - \sqrt{3} < a < -1 + \sqrt{3}$.

Since we chose $a < -1$, the intersection is $-1 - \sqrt{3} < a < -1$.

However, since $a = -1 - \sqrt{3}$ is a root of the equation $a^2 + 2a - 2 = 0$, and for $a < -1$ we require $(a + 1)^2 < 3$, the values of $a$ that satisfy are $a < -1 - \sqrt{3}$.

Therefore, the values of $a$ that satisfy are $a < -1 - \sqrt{3}$.