Question

what is the solution to the equation?
\\(\sqrt{x} + 2 = x\\)
\\(\bigcirc\\) a. \\(1\\)
\\(\bigcirc\\) b. \\(4\\)
\\(\bigcirc\\) c. \\(1\\) and \\(4\\)
\\(\bigcirc\\) d. no solution

Explanation:

Step1: Isolate the square root

Subtract 2 from both sides to get $\sqrt{x} = x - 2$.

Step2: Square both sides

Square both sides to eliminate the square root: $(\sqrt{x})^2 = (x - 2)^2$, which simplifies to $x = x^2 - 4x + 4$.

Step3: Rearrange into quadratic equation

Rearrange the equation to standard quadratic form: $x^2 - 5x + 4 = 0$.

Step4: Factor the quadratic

Factor the quadratic: $(x - 1)(x - 4) = 0$. So the solutions from factoring are $x = 1$ and $x = 4$.

Step5: Check for extraneous solutions

• For $x = 1$: Substitute into the original equation $\sqrt{1} + 2 = 1$ → $1 + 2 = 1$ → $3 = 1$, which is false.
• For $x = 4$: Substitute into the original equation $\sqrt{4} + 2 = 4$ → $2 + 2 = 4$ → $4 = 4$, which is true. But wait, when we squared both sides, we might have introduced an extraneous solution. Wait, but let's re - check the isolation step. Wait, originally the equation is $\sqrt{x}+2 = x$. When we isolate $\sqrt{x}$, we get $\sqrt{x}=x - 2$. But the square root function $\sqrt{x}$ has a range of $y\geq0$, so $x - 2\geq0$ → $x\geq2$. For $x = 1$, it doesn't satisfy $x\geq2$, so it's extraneous. For $x = 4$, it satisfies $x\geq2$ and the original equation. Wait, but let's check again. Wait, when $x = 4$, $\sqrt{4}+2=2 + 2=4$, which is correct. But wait, the option B is 4, but let's check the initial isolation. Wait, maybe I made a mistake in the check for $x = 1$. Wait, $\sqrt{1}+2=1 + 2 = 3

eq1$, so $x = 1$ is extraneous. $x = 4$: $\sqrt{4}+2=2 + 2 = 4$, which is correct. But wait, the options are A.1, B.4, C.1 and 4, D. no solution. Wait, but when we solved, $x = 4$ is a valid solution? Wait, no, wait the original equation is $\sqrt{x}+2=x$. Let's graph the functions $y=\sqrt{x}+2$ and $y = x$. The square root function shifted up by 2. The line $y = x$ has a slope of 1. The intersection: when $x = 4$, $y = 4$ and $y=\sqrt{4}+2 = 4$. When $x = 1$, $y=\sqrt{1}+2 = 3
eq1$. So $x = 4$ is a solution? Wait, but let's check the step where we isolated the square root. $\sqrt{x}=x - 2$. The left - hand side $\sqrt{x}\geq0$, so the right - hand side $x - 2\geq0$ → $x\geq2$. So $x = 1$ is less than 2, so it's extraneous. $x = 4$ is greater than 2, so it's a valid solution. Wait, but the option B is 4. But wait, let's re - do the check. $\sqrt{4}+2=2 + 2 = 4$, which is equal to $x = 4$. So $x = 4$ is a solution. But wait, the answer options: B is 4. But wait, maybe I made a mistake in the quadratic solution. Wait, $x^2-5x + 4=(x - 1)(x - 4)=0$, so roots at 1 and 4. After checking, only $x = 4$ is valid. So the solution is $x = 4$, which is option B. Wait, but earlier when I thought about the range, $\sqrt{x}=x - 2$, so $x-2\geq0$ → $x\geq2$. So $x = 4$ is in the domain, and it works. $x = 1$ is not in the domain of the isolated square root, so it's extraneous. So the solution is $x = 4$.

Answer:

B. 4