Question

Determine the set of all real numbers for this inequality:
(x-1)^2<=lxl

Answer 1

`(-√(5)+3 )/(2) \leq x\leq (√(5)+3 )/(2)`

Explanation:

Alright, let's tackle this inequality step by step. The absolute value, denoted by
`\(|x|\)`, is defined as follows:

`\[ |x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases} \]`

Now, the given inequality is
`\((x-1)^2 \leq |x|\)`. We can break this into cases based on the sign of x:

1. When
`\(x \geq 0\)`: In this case,
`\(|x| = x\)`, so the inequality becomes
`\((x-1)^2 \leq x\)`.

2. When
`\(x < 0\)`: In this case,
`\(|x| = -x\)`, so the inequality becomes
`\((x-1)^2 \leq -x\)`.

Now, let's solve each case separately:

1.
`\(x \geq 0\)`:

`\[ (x-1)^2 \leq x \]`

Expand and simplify:

`\[ x^2 - 2x + 1 \leq x \]`

Rearrange:

`\[ x^2 - 3x + 1 \leq 0 \]`

Solve for x: This quadratic inequality holds when x is within the interval
`\(\left[ (3 - √(5))/(2), (3 + √(5))/(2) \right]\)`.

2.
`\(x < 0\)`:

`\[ (x-1)^2 \leq -x \]`

Expand and simplify:

`\[ x^2 - 2x + 1 \leq -x \]`

Rearrange:

`\[ x^2 - 3x + 1 \leq 0 \]`

Interestingly, the inequality is the same as in case 1.

So, the solution to the given inequality is the set of all real numbers x within the interval
`\(\left[ (3 - √(5))/(2), (3 + √(5))/(2) \right]\)`.