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

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Discussion Overview

The discussion revolves around solving the quadratic inequality x - [10/(x - 1)] ≥ 4. Participants explore the steps involved in manipulating the inequality and determining the solution set, focusing on the algebraic transformations and testing intervals.

Discussion Character

  • Homework-related
  • Mathematical reasoning

Main Points Raised

  • One participant suggests starting by subtracting 4 from both sides and simplifying the left-hand side.
  • Another participant provides a transformed version of the inequality, stating it simplifies to (x + 1)(x - 6)/(x - 1) ≥ 0 and notes the exclusion of x = 1 due to division by zero.
  • A participant mentions testing various intervals to determine where the inequality holds true, including critical values of x such as -1 and 6.
  • Participants discuss the inclusion of critical points in the solution set, with one participant concluding the solution as [-1, 1) U [6, infinity).

Areas of Agreement / Disagreement

Participants generally agree on the steps to manipulate the inequality and the critical points to consider, but there is no explicit consensus on the final presentation of the solution or the interpretation of the intervals.

Contextual Notes

Some assumptions regarding the behavior of the inequality at critical points and the treatment of the weak inequality are not fully explored, leaving room for further discussion.

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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?
 
Last edited:
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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)
 
Last edited:

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