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To solve the equation [tex]\(2 \sqrt{x-1} + 2 = \frac{3x}{x-1}\)[/tex], we can approximate the solution by evaluating the expression on both sides of the equation for various values of [tex]\(x\)[/tex]. Let's construct a table of values for the chosen [tex]\(x\)[/tex] values: 2.5, 2.75, 3, and 4.75, and find which one is closest to satisfying the equation.
First, let's plug in each value into the left-hand side [tex]\(2 \sqrt{x-1} + 2\)[/tex] and the right-hand side [tex]\(\frac{3x}{x-1}\)[/tex]:
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]
[tex]\[ \text{RHS: } \frac{3 \cdot 2.5}{2.5 - 1} = \frac{7.5}{1.5} = 5 \][/tex]
[tex]\[ \text{Difference: } 4.4494 - 5 \approx -0.5505 \][/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]
[tex]\[ \text{RHS: } \frac{3 \cdot 2.75}{2.75 - 1} = \frac{8.25}{1.75} \approx 4.7143 \][/tex]
[tex]\[ \text{Difference: } 4.6458 - 4.7143 \approx -0.0685 \][/tex]
3. For [tex]\(x = 3\)[/tex]:
[tex]\[ \text{LHS: } 2 \sqrt{3-1} + 2 = 2 \sqrt{2} + 2 \approx 2 \times 1.4142 + 2 = 2.8284 + 2 = 4.8284 \][/tex]
[tex]\[ \text{RHS: } \frac{3 \cdot 3}{3 - 1} = \frac{9}{2} = 4.5 \][/tex]
[tex]\[ \text{Difference: } 4.8284 - 4.5 \approx 0.3284 \][/tex]
4. For [tex]\(x = 4.75\)[/tex]:
[tex]\[ \text{LHS: } 2 \sqrt{4.75-1} + 2 = 2 \sqrt{3.75} + 2 \approx 2 \times 1.9365 + 2 = 3.8730 + 2 = 5.8730 \][/tex]
[tex]\[ \text{RHS: } \frac{3 \cdot 4.75}{4.75 - 1} = \frac{14.25}{3.75} \approx 3.80 \][/tex]
[tex]\[ \text{Difference: } 5.8730 - 3.80 = 2.073 \][/tex]
From these calculations, we observe that the smallest difference in magnitude is for [tex]\(x = 2.75\)[/tex], with a value of approximately [tex]\(-0.0685\)[/tex].
Therefore, the closest approximation to the solution of the equation [tex]\(2 \sqrt{x-1} + 2 = \frac{3x}{x-1}\)[/tex] to the nearest fourth of a unit is:
B. [tex]\(x \approx 2.75\)[/tex]
First, let's plug in each value into the left-hand side [tex]\(2 \sqrt{x-1} + 2\)[/tex] and the right-hand side [tex]\(\frac{3x}{x-1}\)[/tex]:
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]
[tex]\[ \text{RHS: } \frac{3 \cdot 2.5}{2.5 - 1} = \frac{7.5}{1.5} = 5 \][/tex]
[tex]\[ \text{Difference: } 4.4494 - 5 \approx -0.5505 \][/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]
[tex]\[ \text{RHS: } \frac{3 \cdot 2.75}{2.75 - 1} = \frac{8.25}{1.75} \approx 4.7143 \][/tex]
[tex]\[ \text{Difference: } 4.6458 - 4.7143 \approx -0.0685 \][/tex]
3. For [tex]\(x = 3\)[/tex]:
[tex]\[ \text{LHS: } 2 \sqrt{3-1} + 2 = 2 \sqrt{2} + 2 \approx 2 \times 1.4142 + 2 = 2.8284 + 2 = 4.8284 \][/tex]
[tex]\[ \text{RHS: } \frac{3 \cdot 3}{3 - 1} = \frac{9}{2} = 4.5 \][/tex]
[tex]\[ \text{Difference: } 4.8284 - 4.5 \approx 0.3284 \][/tex]
4. For [tex]\(x = 4.75\)[/tex]:
[tex]\[ \text{LHS: } 2 \sqrt{4.75-1} + 2 = 2 \sqrt{3.75} + 2 \approx 2 \times 1.9365 + 2 = 3.8730 + 2 = 5.8730 \][/tex]
[tex]\[ \text{RHS: } \frac{3 \cdot 4.75}{4.75 - 1} = \frac{14.25}{3.75} \approx 3.80 \][/tex]
[tex]\[ \text{Difference: } 5.8730 - 3.80 = 2.073 \][/tex]
From these calculations, we observe that the smallest difference in magnitude is for [tex]\(x = 2.75\)[/tex], with a value of approximately [tex]\(-0.0685\)[/tex].
Therefore, the closest approximation to the solution of the equation [tex]\(2 \sqrt{x-1} + 2 = \frac{3x}{x-1}\)[/tex] to the nearest fourth of a unit is:
B. [tex]\(x \approx 2.75\)[/tex]
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