Question 186·Hard·Nonlinear Equations in One Variable; Systems in Two Variables

A square root is never negative. Since it equals $x-1$, the right side must be nonnegative: $x-1\ge0$. Add $1$ to both sides: $x\ge1$. Any value below $1$ cannot satisfy the original system.

Type both equations as given into Desmos. Click where the graphs meet. Desmos shows approximately $(4.449,3.449)$. An intersection lies on both graphs, so the first coordinate tells you that the required $x$ is about $4.449$.

Type the four choices on new Desmos lines. Desmos prints about $4.449$, $-0.449$, $3.449$, and $1.449$, in that order. Only $2+\sqrt6$ matches the $x$-coordinate, not the $y$-coordinate, of the intersection. So the value of $x$ is $2+\sqrt6$. Choice A.

Both equations give $y$, so at a solution their right sides must be equal: $\sqrt{2x+3}=x-1$. Now there's only one unknown to find.

Square both whole sides: $2x+3=(x-1)^2$. Squaring can erase a sign difference: positive and negative numbers can have the same square. So any values this equation gives still need to pass the original equation.

Expand the square: $2x+3=x^2-2x+1$. Subtract $2x$ from both sides: $3=x^2-4x+1$. Subtract $3$ from both sides: $0=x^2-4x-2$. This quadratic equation has an $x^2$ term, so it may give two candidates.

Half of $-4$ is $-2$, and its square is $4$. Adding $4$ will turn $x^2-4x$ into one squared expression. First add $2$ to both sides: $x^2-4x=2$. Add $4$ to both sides: $x^2-4x+4=6$. Rewrite the left side: $(x-2)^2=6$.

Taking square roots of a squared expression gives both signs, since either a positive or a negative number can square to $6$: $x-2=\pm\sqrt6$. Add $2$ to both sides: $x=2\pm\sqrt6$.

For $x=2-\sqrt6$, the right side of the original equation is $x-1=1-\sqrt6<0$. It cannot equal a square root. The other candidate has $x-1=1+\sqrt6>0$, so it passes the sign check and the squared equation. The value of $x$ that satisfies the system is $2+\sqrt6$. Choice A.