Question

which statement describes how to solve \\(\sqrt{3x + 4} = \sqrt{3x} + 4\\)?
\\(\circ\\) square both sides once and then solve the resulting linear equation.
\\(\circ\\) square both sides once and then solve the resulting quadratic equation.
\\(\circ\\) square both sides twice and then solve the resulting linear equation.
\\(\circ\\) square both sides twice and then solve the resulting quadratic equation.

Explanation:

Step1: Analyze the equation

The equation is \(\sqrt{3x + 4}=-\sqrt{3x}+4\). To eliminate the square roots, we first square both sides.
Left side squared: \((\sqrt{3x + 4})^2 = 3x + 4\)
Right side squared: \((-\sqrt{3x}+4)^2=(-\sqrt{3x})^2 + 2\times(-\sqrt{3x})\times4+4^2 = 3x-8\sqrt{3x}+16\)
So after first squaring, we have \(3x + 4=3x-8\sqrt{3x}+16\). Simplify this:
Subtract \(3x\) from both sides: \(4=-8\sqrt{3x}+16\)
Then, we still have a square root (\(\sqrt{3x}\)) in the equation. So we need to isolate the square root term:
\(- 8\sqrt{3x}=4 - 16=-12\)
\(\sqrt{3x}=\frac{- 12}{-8}=\frac{3}{2}\)
Wait, no, actually when we simplify \(3x + 4=3x-8\sqrt{3x}+16\), subtract \(3x\) from both sides: \(4=-8\sqrt{3x}+16\), then subtract 16: \(4 - 16=-8\sqrt{3x}\), \(-12=-8\sqrt{3x}\), then \(\sqrt{3x}=\frac{12}{8}=\frac{3}{2}\). But if we didn't notice that, we would see that after first squaring, we still have a square root, so we need to square both sides again? Wait, no, in the above case, after first squaring, we can isolate the square root term. Wait, maybe my initial analysis was wrong. Wait the original equation is \(\sqrt{3x + 4}=-\sqrt{3x}+4\). Let's do it again.

Wait, maybe a better approach: Let's denote \(y = \sqrt{3x}\), then the equation becomes \(\sqrt{y^{2}+4}=-y + 4\). Square both sides: \(y^{2}+4=y^{2}-8y + 16\). Subtract \(y^{2}\): \(4=-8y + 16\), then \(8y=12\), \(y=\frac{3}{2}\). Then \(\sqrt{3x}=\frac{3}{2}\), square both sides: \(3x=\frac{9}{4}\), \(x = \frac{3}{4}\). But in terms of the process of solving the equation \(\sqrt{3x + 4}=-\sqrt{3x}+4\), when we square both sides the first time, we get an equation with a remaining square root? Wait no, in the step where we squared \((-\sqrt{3x}+4)^2\), we have a square root term, but when we simplify, we can isolate it. But actually, if we consider the general process, when we have an equation with square roots, if after first squaring, there is still a square root, we need to square again. Wait, in the given equation, after first squaring, we get \(3x + 4=3x-8\sqrt{3x}+16\), which simplifies to \(4=-8\sqrt{3x}+16\), then we can solve for \(\sqrt{3x}\) (which is a linear equation in terms of \(\sqrt{3x}\)), but if we didn't simplify, we would think we need to square again. Wait, maybe the question is about the process. Wait the options:

Option 1: Square once, solve linear. But after first square, we have a square root left? Wait no, in our simplification, after first square, we can get rid of the square root by simplifying. Wait, maybe I made a mistake. Let's re - express the equation after first squaring:

Original equation: \(\sqrt{3x + 4}=-\sqrt{3x}+4\)

Square both sides:

LHS: \(3x + 4\)

RHS: \((-\sqrt{3x}+4)^2=3x-8\sqrt{3x}+16\)

So equation: \(3x + 4=3x-8\sqrt{3x}+16\)

Subtract \(3x\) from both sides: \(4=-8\sqrt{3x}+16\)

Now, this is a linear equation in terms of \(\sqrt{3x}\). Let's let \(z = \sqrt{3x}\), then \(4=-8z + 16\), which is a linear equation (\(8z=16 - 4 = 12\), \(z=\frac{12}{8}=\frac{3}{2}\)). Then we can solve for \(x\) by squaring \(z=\frac{3}{2}\) (i.e., \(\sqrt{3x}=\frac{3}{2}\), square both sides: \(3x=\frac{9}{4}\), \(x=\frac{3}{4}\)). But the process of solving the equation after first squaring leads to a linear equation in the square root term. Wait, but the options are about the number of times we square and the type of equation after squaring.

Wait, maybe the question is designed to consider that when we have two square roots or a square root and a linear term, sometimes we need to square twice. Wait, in our case, after first squaring, we still have a square root? No, in the equation \(4=-8\sqrt{3x}+16\), the square root is still there, so we need to isolate the square root and square again? Wait, no, in this equation, we can solve for \(\sqrt{3x}\) without squaring again. Let's see:

\(4=-8\sqrt{3x}+16\)

Subtract 16: \(4 - 16=-8\sqrt{3x}\)

\(-12=-8\sqrt{3x}\)

Divide by - 8: \(\sqrt{3x}=\frac{12}{8}=\frac{3}{2}\)

Then square both sides: \(3x=\frac{9}{4}\), \(x=\frac{3}{4}\)

So in total, we squared twice? Wait, first square the original equation, then square the equation \(\sqrt{3x}=\frac{3}{2}\). So we squared both sides twice. And after the second squaring, we solve a linear equation (for \(x\))? Wait, no, when we square \(\sqrt{3x}=\frac{3}{2}\), we get \(3x=\frac{9}{4}\), which is a linear equation. Wait, but the options:

Option C: Square both sides twice and then solve the resulting linear equation.

Wait, let's re - examine the process:

1. Start with \(\sqrt{3x + 4}=-\sqrt{3x}+4\)

2. Square both sides (first time): \(3x + 4 = 3x-8\sqrt{3x}+16\) (this still has a square root)

3. Simplify to isolate the square root: \(-8\sqrt{3x}=4 - 16=-12\) → \(\sqrt{3x}=\frac{3}{2}\) (now we have a square root equation, so we square both sides again (second time))

4. Square both sides (second time): \(3x=\frac{9}{4}\) (this is a linear equation)

5. Solve the linear equation: \(x=\frac{3}{4}\)

So we squared both sides twice, and then solved a linear equation. So the correct option is "Square both sides twice and then solve the resulting linear equation." (the third option)

Answer:

C. Square both sides twice and then solve the resulting linear equation.