A square root like $\sqrt{2x+3}$ is always nonnegative. What must be true about $x - 1$ for the equation $\sqrt{2x+3} = x - 1$ to hold?

After you set $\sqrt{2x+3} = x - 1$, square both sides to remove the square root, then rearrange to get a quadratic equation.

Solving the quadratic will give two values of $x$. Plug them back into the original equation (or at least check the sign condition you found) to see which one really works.

In Desmos, enter y = sqrt(2x + 3) on one line and y = x - 1 on another. Make sure the square root is typed as sqrt(2x + 3).

Look for the point where the two graphs intersect. Tap or click on the intersection to see its coordinates; note the x-coordinate of this point.

Compare the x-coordinate from Desmos with the answer choices by approximating values like $2 + \sqrt{6}$ and $2 - \sqrt{6}$ (you can type sqrt(6) in Desmos to get a decimal) and see which choice equals the intersection x-value.

Both equations equal $y$, so set their right-hand sides equal:

This is the equation you need to solve for $x$.

A square root always gives a nonnegative result. That means the right-hand side must be at least $0$:

• Since $\sqrt{2x+3} \ge 0$, we must have $x - 1 \ge 0$.
• So $x \ge 1$.

Also, the expression under the square root must be nonnegative:

• $2x + 3 \ge 0$ gives $x \ge -\tfrac{3}{2}$.

Together, the stricter condition is $x \ge 1$. Any final solution must satisfy this.

To solve $\sqrt{2x+3} = x - 1$, square both sides:

so

Now bring everything to one side to form a quadratic:

Solve $x^2 - 4x - 2 = 0$ using the quadratic formula.

For $ax^2 + bx + c = 0$ with $a = 1$, $b = -4$, $c = -2$:

Compute inside the square root:

So

Algebraically, there are two possible solutions: $x = 2 + \sqrt{6}$ and $x = 2 - \sqrt{6}$.

You must check both $x = 2 + \sqrt{6}$ and $x = 2 - \sqrt{6}$ in the original equation $\sqrt{2x+3} = x - 1$ and against the condition $x \ge 1$.

• Approximate $\sqrt{6} \approx 2.45$.
  • $2 + \sqrt{6} \approx 4.45$, which is $\ge 1$, so it could be valid.
  • $2 - \sqrt{6} \approx -0.45$, which is less than $1$, so it cannot satisfy $x - 1 \ge 0$.

Therefore $x = 2 - \sqrt{6}$ is an extraneous solution created by squaring, and the only value of $x$ that satisfies the system is $\boxed{2+\sqrt{6}}$.