MHB How Do You Solve This Quadratic Inequality: x - [10/(x - 1)] ≥ 4?

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To solve the quadratic inequality x - [10/(x - 1)] ≥ 4, start by rearranging the equation to isolate the terms, resulting in (x + 1)(x - 6)/(x - 1) ≥ 0. Exclude x = 1 due to division by zero and test intervals around the critical points -1, 1, and 6. The valid solutions are found to be in the intervals [-1, 1) and [6, ∞). The final solution confirms the inclusion of critical values where the inequality holds true.
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Section 2.6
Question 68Solve the quadratic inequality.

x - [10/(x - 1)] ≥ 4

I begin by subtracting 4 from both sides and then simplify the left hand side, right?
 
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RTCNTC said:
Section 2.6
Question 68Solve the quadratic inequality.

x - [10/(x - 1)] ≥ 4

I begin by subtracting 4 from both sides and then simplify the left hand side, right?

Yes, you should wind up with:

$$\frac{(x+1)(x-6)}{x-1}\ge0$$

And then observing we have a weak inequality, and that $x=1$ must be excluded (division by zero), testing 1 interval, and allowing the others to alternate signs, we wind up with:

$$[-1,1)\,\cup\,[6,\infty)$$
 
I will show my work tomorrow.
 
(x + 1)(x - 6)/(x - 1) ≥ 0

We exclude x = 1 because it creates (expression/0) but include it on the number line.

<-------(-1)----------(1)---------(6)--------->

If we let x = -1 and 6, the inequality becomes 0 ≥ 0, which is true. So, we include our critical values of x.

For (-infinity, -1), let x = -2. False statement.

For (-1, 1), let x = 0. True statement.

For (1, 6), let x = 2. False statement.

For (6, infinity), let x = 7. True statement.

Solution:

[-1, 1) U [6, infinity)
 
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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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