Question

directions: use your answers to determine w letter of the problem at each station in the c
b
\sqrt{3x + 7} - x = 3

Explanation:

Step1: Isolate the square root

Add \( x \) to both sides of the equation \( \sqrt{3x + 7}-x = 3 \) to get \( \sqrt{3x + 7}=x + 3 \).

Step2: Square both sides

Square both sides of the equation \( \sqrt{3x + 7}=x + 3 \) to eliminate the square root. Using the formula \( (a + b)^2=a^2+2ab + b^2 \), we have \( (\sqrt{3x + 7})^2=(x + 3)^2 \), which simplifies to \( 3x+7=x^{2}+6x + 9 \).

Step3: Rearrange into quadratic form

Rearrange the equation \( 3x + 7=x^{2}+6x + 9 \) to get a quadratic equation in standard form \( ax^{2}+bx + c = 0 \). Subtract \( 3x \) and \( 7 \) from both sides: \( x^{2}+6x + 9-3x - 7 = 0 \), which simplifies to \( x^{2}+3x + 2 = 0 \).

Step4: Factor the quadratic equation

Factor the quadratic equation \( x^{2}+3x + 2 = 0 \). We need two numbers that multiply to \( 2 \) and add up to \( 3 \). These numbers are \( 1 \) and \( 2 \), so the equation factors as \( (x + 1)(x + 2)=0 \).

Step5: Solve for \( x \)

Set each factor equal to zero and solve for \( x \). For \( x + 1 = 0 \), we get \( x=-1 \). For \( x + 2 = 0 \), we get \( x=-2 \).

Step6: Check for extraneous solutions

We need to check these solutions in the original equation because squaring both sides can introduce extraneous solutions.
- For \( x=-1 \): Substitute into the original equation \( \sqrt{3(-1)+7}-(-1)=\sqrt{-3 + 7}+1=\sqrt{4}+1 = 2 + 1=3 \). This satisfies the original equation.
- For \( x=-2 \): Substitute into the original equation \( \sqrt{3(-2)+7}-(-2)=\sqrt{-6 + 7}+2=\sqrt{1}+2 = 1 + 2 = 3 \). Wait, that also seems to satisfy? Wait, maybe I made a mistake in factoring or calculation. Wait, let's re - check the squaring step.

Wait, original equation: \( \sqrt{3x + 7}-x = 3 \). When \( x=-2 \), left - hand side: \( \sqrt{3\times(-2)+7}-(-2)=\sqrt{1}+2 = 3 \), right - hand side is \( 3 \). When \( x=-1 \), left - hand side: \( \sqrt{3\times(-1)+7}-(-1)=\sqrt{4}+1 = 3 \), right - hand side is \( 3 \). But wait, let's check the domain of the square root. The expression under the square root \( 3x + 7\geq0\), so \( x\geq-\frac{7}{3}\approx - 2.333\). Both \( x=-1 \) and \( x=-2 \) are in the domain. But wait, maybe there is a miscalculation in the problem - solving process. Wait, let's go back to the original equation: \( \sqrt{3x + 7}-x = 3 \), or \( \sqrt{3x + 7}=x + 3 \). For the right - hand side \( x + 3\), since the square root is non - negative, \( x+3\geq0\), so \( x\geq - 3 \).

Wait, maybe the problem was written as \( \sqrt{3x + 7}-x = 3 \) (maybe a typo in the original image, maybe it's \( \sqrt{3x + 7}-x = 3 \) or \( \sqrt{3x + 7}+x = 3 \)? Wait, the user's image shows \( \sqrt{3x + 7}-x = 3 \). But according to the calculation, both \( x=-1 \) and \( x=-2 \) are solutions. But maybe I made a mistake. Wait, let's re - do the quadratic equation:

From \( \sqrt{3x + 7}=x + 3 \), squaring both sides gives \( 3x + 7=x^{2}+6x + 9 \), then \( x^{2}+3x + 2 = 0 \), factored as \( (x + 1)(x + 2)=0 \), so \( x=-1 \) or \( x=-2 \).

But let's check again:

For \( x=-2 \):

\( \sqrt{3\times(-2)+7}-(-2)=\sqrt{1}+2 = 3 \), which is equal to the right - hand side.

For \( x=-1 \):

\( \sqrt{3\times(-1)+7}-(-1)=\sqrt{4}+1 = 3 \), which is also equal to the right - hand side.

But maybe the original problem was \( \sqrt{3x + 7}+x = 3 \)? If that's the case, the solution would be different. Assuming the problem is \( \sqrt{3x + 7}-x = 3 \), the solutions are \( x=-1 \) and \( x=-2 \). But maybe there is a mistake in my process.

Wait, let's start over:

Given \( \sqrt{3x + 7}-x = 3 \)

1. Isolate the square root: \( \sqrt{3x + 7}=x + 3 \)
2. Square both sides: \( 3x + 7=(x + 3)^2=x^{2}+6x + 9 \)
3. Rearrange: \( x^{2}+3x + 2 = 0 \)
4. Factor: \( (x + 1)(x + 2)=0 \)
5. Solutions: \( x=-1 \) or \( x=-2 \)

Both solutions satisfy the original equation. But maybe the problem was intended to have only one solution, so perhaps there was a typo in the problem statement. If we assume that the original equation is \( \sqrt{3x + 7}+x = 3 \), then:

1. Isolate the square root: \( \sqrt{3x + 7}=3 - x \)
2. Square both sides: \( 3x + 7 = 9-6x+x^{2} \)
3. Rearrange: \( x^{2}-9x + 2 = 0 \)
4. Use quadratic formula \( x=\frac{9\pm\sqrt{81 - 8}}{2}=\frac{9\pm\sqrt{73}}{2}\), but this is more complicated.

Given that the user's problem is \( \sqrt{3x + 7}-x = 3 \), the solutions are \( x=-1 \) and \( x=-2 \). But maybe in the original problem, there was a mistake, and the correct solution is \( x = 1 \) or \( x = 2 \)? Wait, no, let's check with \( x = 1 \): \( \sqrt{3\times1+7}-1=\sqrt{10}-1\approx3.16 - 1 = 2.16 eq3 \). With \( x = 2 \): \( \sqrt{3\times2+7}-2=\sqrt{13}-2\approx3.606 - 2 = 1.606 eq3 \).

Wait, I think I made a mistake in the sign when moving terms. Let's re - do the equation:

Original equation: \( \sqrt{3x + 7}-x = 3 \)

Add \( x \) to both sides: \( \sqrt{3x + 7}=x + 3 \)

Square both sides: \( 3x+7=x^{2}+6x + 9 \)

Subtract \( 3x \) and \( 7 \) from both sides: \( 0=x^{2}+6x + 9-3x - 7 \)

So \( x^{2}+3x + 2 = 0 \), which factors as \( (x + 1)(x + 2)=0 \), so \( x=-1 \) or \( x=-2 \). And both satisfy the original equation as we saw.

Answer:

The solutions of the equation \( \sqrt{3x + 7}-x = 3 \) are \( x=-1 \) and \( x=-2 \). If we consider that maybe there was a typo and the equation is \( \sqrt{3x + 7}+x = 3 \), the solution process would be different. But based on the given equation \( \sqrt{3x + 7}-x = 3 \), the solutions are \( x=-1 \) and \( x=-2 \).