To find the approximate solution to the equation [tex]\(2 \sqrt{x - 1} + 2 = \frac{3x}{x - 1}\)[/tex] to the nearest fourth of a unit, we can use a table of values to evaluate the left-hand side (LHS) and right-hand side (RHS) of the equation at different values of [tex]\(x\)[/tex] and find where they are approximately equal.
Let's create a table for the given possible choices [tex]\(x = 2.5\)[/tex], [tex]\(x = 2.75\)[/tex], [tex]\(x = 3\)[/tex], and [tex]\(x = 4.75\)[/tex].
First, let's evaluate the left-hand side [tex]\(2 \sqrt{x - 1} + 2\)[/tex] for each value:
1. For [tex]\(x = 2.5\)[/tex]:
[tex]\[
\text{LHS} = 2 \sqrt{2.5 - 1} + 2 = 2 \sqrt{1.5} + 2 \approx 2 \times 1.2247 + 2 \approx 2.4494 + 2 = 4.4494
\][/tex]
2. For [tex]\(x = 2.75\)[/tex]:
[tex]\[
\text{LHS} = 2 \sqrt{2.75 - 1} + 2 = 2 \sqrt{1.75} + 2 \approx 2 \times 1.3229 + 2 \approx 2.6458 + 2 = 4.6458
\][/tex]
3. For [tex]\(x = 3\)[/tex]:
[tex]\[
\text{LHS} = 2 \sqrt{3 - 1} + 2 = 2 \sqrt{2} + 2 \approx 2 \times 1.4142 + 2 \approx 2.8284 + 2 = 4.8284
\][/tex]
4. For [tex]\(x = 4.75\)[/tex]:
[tex]\[
\text{LHS} = 2 \sqrt{4.75 - 1} + 2 = 2 \sqrt{3.75} + 2 \approx 2 \times 1.9365 + 2 \approx 3.8730 + 2 = 5.8730
\][/tex]
Next, let's evaluate the right-hand side [tex]\(\frac{3x}{x-1}\)[/tex] for each value:
1. For [tex]\(x = 2.5\)[/tex]:
[tex]\[
\text{RHS} = \frac{3 \times 2.5}{2.5 - 1} = \frac{7.5}{1.5} = 5
\][/tex]
2. For [tex]\(x = 2.75\)[/tex]:
[tex]\[
\text{RHS} = \frac{3 \times 2.75}{2.75 - 1} = \frac{8.25}{1.75} \approx 4.7143
\][/tex]
3. For [tex]\(x = 3\)[/tex]:
[tex]\[
\text{RHS} = \frac{3 \times 3}{3 - 1} = \frac{9}{2} = 4.5
\][/tex]
4. For [tex]\(x = 4.75\)[/tex]:
[tex]\[
\text{RHS} = \frac{3 \times 4.75}{4.75 - 1} = \frac{14.25}{3.75} \approx 3.8
\][/tex]
Now, let's compare LHS and RHS:
1. For [tex]\(x = 2.5\)[/tex]:
[tex]\[
\text{LHS} = 4.4494, \quad \text{RHS} = 5 \quad \Rightarrow \quad \text{not close}
\][/tex]
2. For [tex]\(x = 2.75\)[/tex]:
[tex]\[
\text{LHS} = 4.6458, \quad \text{RHS} = 4.7143 \quad \Rightarrow \quad \text{very close}
\][/tex]
3. For [tex]\(x = 3\)[/tex]:
[tex]\[
\text{LHS} = 4.8284, \quad \text{RHS} = 4.5 \quad \Rightarrow \quad \text{not close}
\][/tex]
4. For [tex]\(x = 4.75\)[/tex]:
[tex]\[
\text{LHS} = 5.8730, \quad \text{RHS} = 3.8 \quad \Rightarrow \quad \text{not close}
\][/tex]
The closest match, where LHS and RHS are approximately equal, occurs for [tex]\(x = 2.75\)[/tex].
Thus, the correct answer is:
A. [tex]\( x \approx 2.75 \)[/tex]
