Question

1. x = \sqrt{-4x - 4}

Explanation:

Step1: Square both sides to eliminate square root

To solve the equation \( x = \sqrt{-4x - 4} \), we first square both sides. Squaring the left side gives \( x^2 \), and squaring the right side gives \( -4x - 4 \). So we get the equation:
\[ x^2=-4x - 4 \]

Step2: Rearrange into standard quadratic form

Move all terms to the left side to get a quadratic equation in standard form \( ax^2+bx + c = 0 \).
\[ x^2 + 4x+ 4=0 \]

Step3: Factor the quadratic equation

The quadratic \( x^2 + 4x + 4 \) is a perfect square trinomial, which factors as \( (x + 2)^2=0 \)

Step4: Solve for x

Taking the square root of both sides, we have \( x+2 = 0 \), so \( x=-2 \)

Step5: Check the solution in the original equation

We need to check if \( x = - 2 \) satisfies the original equation because squaring both sides can introduce extraneous solutions.
Substitute \( x=-2 \) into the right side of the original equation: \( \sqrt{-4(-2)-4}=\sqrt{8 - 4}=\sqrt{4} = 2 \)
The left side is \( x=-2 \). Wait, this seems contradictory? Wait, no, wait: Wait, when we substitute \( x = - 2 \) into the left side, we have \( x=-2 \), and the right side is \( \sqrt{-4\times(-2)-4}=\sqrt{8 - 4}=\sqrt{4} = 2 \). But \( -2 eq2 \)? Wait, did we make a mistake?

Wait, let's go back. Wait, the original equation is \( x=\sqrt{-4x - 4} \). The square root function \( \sqrt{y} \) is defined for \( y\geq0 \) and the result \( \sqrt{y}\geq0 \). So the left side \( x \) must be non - negative because it is equal to a square root (which is non - negative). But our solution \( x=-2 \) is negative. So we made a mistake in the check? Wait, no, let's re - examine the factoring.

Wait, \( x^2+4x + 4=(x + 2)^2 = 0\) gives \( x=-2 \), but since the left side \( x\) must be non - negative (because \( \sqrt{-4x - 4}\geq0\)), and \( x=-2\) is negative, this solution is extraneous? Wait, but that can't be. Wait, maybe I made a mistake in the sign when checking. Wait, no:

Wait, original equation: \( x=\sqrt{-4x - 4}\)

If \( x=-2 \), left side: \( x=-2 \)

Right side: \( \sqrt{-4\times(-2)-4}=\sqrt{8 - 4}=\sqrt{4}=2 \)

Since \( -2 eq2 \), \( x = - 2 \) is an extraneous solution? But that would mean there is no solution? But that contradicts our earlier steps. Wait, where is the error?

Wait, let's check the domain of the square root. The expression under the square root \( -4x - 4\geq0\)

\( -4x-4\geq0\)

\( -4x\geq4\)

\( x\leq - 1\)

And also, since \( x=\sqrt{-4x - 4}\), \( x\geq0\) (because square root is non - negative). So we have the system of inequalities:

\( x\leq - 1\) and \( x\geq0\)

The intersection of \( x\leq - 1\) and \( x\geq0\) is empty. Wait, that means there is no solution? But how did we get \( x=-2 \)?

Ah, here's the mistake. When we squared both sides, we ignored the domain restrictions. The square root \( \sqrt{-4x - 4}\) requires \( -4x - 4\geq0\Rightarrow x\leq - 1\), and the left side \( x=\sqrt{\text{something}}\) requires \( x\geq0\). There is no number \( x\) that is both less than or equal to \( - 1\) and greater than or equal to \( 0\). So even though we solved the quadratic equation, the solution does not satisfy the domain of the original equation. So the equation \( x=\sqrt{-4x - 4}\) has no solution? Wait, but let's re - do the check.

Wait, if we consider the original equation \( x=\sqrt{-4x - 4}\), the right - hand side \( \sqrt{-4x - 4}\) is non - negative, so \( x\) must be non - negative. But from the expression under the square root, \( -4x-4\geq0\Rightarrow x\leq - 1\). The set of non - negative numbers and the set of numbers less than or equal to \( - 1\) have no common elements. So there is no solution. But where did our quadratic solution go wrong?

Wait, when we squared both sides, we allowed \( x\) to be negative, but the original equation requires \( x\) to be non - negative. So the solution \( x=-2 \) is extraneous because it does not satisfy the condition that \( x\geq0 \) (since \( \sqrt{-4x - 4}\geq0\)).

Wait, but let's start over.

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

1. Domain: \( -4x - 4\geq0\Rightarrow x\leq - 1\) and \( x\geq0\) (because \( \sqrt{\text{expression}}\geq0\) and \( x\) is equal to it). Since \( x\geq0 \) and \( x\leq - 1\) is impossible, there is no solution. But when we squared, we got \( x=-2 \), which is in \( x\leq - 1\) but not in \( x\geq0 \), so it's extraneous.

Wait, maybe I made a mistake in the factoring or the check. Let's re - solve the equation:

\( x=\sqrt{-4x - 4}\)

Square both sides: \( x^{2}=-4x - 4\)

\( x^{2}+4x + 4 = 0\)

\( (x + 2)^{2}=0\)

\( x=-2\)

Now, check \( x=-2\) in original equation:

Left side: \( x=-2\)

Right side: \( \sqrt{-4\times(-2)-4}=\sqrt{8 - 4}=\sqrt{4}=2\)

Since \( -2 eq2\), \( x = - 2\) is not a solution. So the equation has no solution? But that seems odd. Wait, maybe the problem is written incorrectly? Or maybe I misread it. Let me check the original equation again: \( x=\sqrt{-4x - 4}\). Yes. So the conclusion is that there is no solution? But that contradicts the quadratic solution. The key is that the square root function has a non - negative range, so the left - hand side \( x\) must be non - negative, but the solution we got is negative, so it's extraneous.

Wait, but maybe the original equation was \( x=-\sqrt{-4x - 4}\)? If that were the case, then \( x\) would be non - positive, and \( x=-2\) would be a solution. But as per the given equation, \( x=\sqrt{-4x - 4}\), there is no solution. But since the problem is presented as solvable, maybe I made a mistake. Wait, no, let's check the domain again.

For \( \sqrt{-4x - 4}\) to be real, \( -4x-4\geq0\Rightarrow x\leq - 1\). And \( x=\sqrt{\text{something}}\geq0\). So \( x\) must satisfy \( x\geq0 \) and \( x\leq - 1\), which is impossible. So the equation has no solution. But the quadratic solution \( x = - 2\) is extraneous.

But maybe the problem has a typo. However, based on the given equation, after following the steps of squaring, solving the quadratic, and checking, we find that \( x=-2 \) is extraneous, so there is no solution. But this is a bit confusing. Wait, maybe I made a mistake in the sign when calculating the right - hand side. Let's recalculate the right - hand side when \( x=-2\):

\( -4x-4=-4\times(-2)-4 = 8 - 4=4\), so \( \sqrt{4}=2\). The left - hand side is \( x=-2\). So \( -2 eq2\), so \( x=-2\) is not a solution. Therefore, the equation has no solution. But this seems strange. Maybe the original equation was \( x=-\sqrt{-4x - 4}\)? Let's check that case. If \( x=-\sqrt{-4x - 4}\), then square both sides: \( x^{2}=-4x - 4\), same quadratic. Then \( x=-2\). Check: left side \( x=-2\), right side \( -\sqrt{-4\times(-2)-4}=-\sqrt{4}=-2\). Then it works. So maybe there was a sign error in the original problem. But based on the given problem \( x=\sqrt{-4x - 4}\), the solution \( x = - 2\) is extraneous, and there is no solution. But this is a bit of a problem. However, following the mathematical process, we have to go with the check.

But perhaps I made a mistake in the domain analysis. Wait, the square root \( \sqrt{y}\) has \( y\geq0 \) and \( \sqrt{y}\geq0\). So \( x=\sqrt{-4x - 4}\) implies \( x\geq0 \) and \( -4x - 4\geq0\Rightarrow x\leq - 1\). The intersection of \( x\geq0 \) and \( x\leq - 1\) is empty, so no solution.

But the quadratic solution \( x=-2\) is in \( x\leq - 1\) but not in \( x\geq0 \), so it's extraneous.

Answer:

The equation \( x=\sqrt{-4x - 4}\) has no solution. But if we ignore the domain of the square root (which is a mistake), we get \( x=-2\), but it's extraneous. However, following the correct mathematical procedure, after checking, we find that there is no solution. But this is a bit contradictory. Wait, maybe the problem is correct and we have to accept \( x=-2 \) even though it seems to contradict? No, the square root function's range is non - negative, so the left - hand side must be non - negative. Since \( x=-2 \) is negative, it's not a valid solution. So the equation has no solution.