Help with inequalities!! Please?

(x - 1)² < 5 → a square cannot be negative, because a square is always positive or null

…but the value inside (…) can be negative or positive.

Example:

(- 2)² = (- 2) * (- 2) = + 4 ← positive value

(+ 2)² = (+ 2) * (+ 2) = + 4 ← positive value

So, in your case, you have 2 possibilities:

First one: (x - 1) is positive → and you must solve: (x - 1) < √5

Second one: (x - 1) is negative → and you must solve: (x - 1) > - √5

If my explanations above are not clear, this is a method:

(x - 1)² < 5

(x - 1)² - 5 < 0

(x - 1)² - (√5)² < 0 → you recognize: a² - b² = (a + b).(a - b)

[(x - 1) + (√5)].[(x - 1) - (√5)] < 0

[x - 1 + √5].[x - 1 - √5] < 0

[x - (1 - √5)].[x - (1 + √5)] < 0

Let: x₁ = 1 - √5

Let: x₂ = 1 + √5

(x - x₁).(x - x₂) < 0 → the roots are: x₁ ; x₂ → then you make a table

x_____-∞_____x₁_____x₂_____+∞

(x - x₁)_____-__0__+______+

(x - x₂)_____-_____-___0___+

sign_______+__0__-___0___+

…and you can see when the sign is < 0

x Є ] x₁ ; x₂ [ → you substitute the real values

→ x Є ] 1 - √5 ; 1 + √5 [

…and this is another method:

(x - 1)² < 5 → the contents of (..) can be positive or negative, so you can write

- 5 < (x - 1)² < 5

- √5 < x - 1 < √5 → you add 1 both sides

- √5 + 1 < x - 1 + 1 < √5 + 1

- √5 + 1 < x < √5 + 1

1 - √5 < x < 1 + √5

→ x Є ] 1 - √5 ; 1 + √5 [

-√5 < |x - 1| < ��5 and since |x - 1| is always ≥ 0 it's actually 0 < |x - 1| < √5

|x - 1| = 0 if x = 1 and |x - 1| = √5 if x = 1 + √5 (about 3.2) or 1 - √5 (about -1.2) so there are these intervals to check:

x < 1 - √5: try x = -2 in original: |-2 - 1| = 3, not < √5 so false

x between 1 - √5 and 1: try x = 0: |0 - 1| < √5 true

x between 1 and 1 + √5 try x = 2: |2 - 1| < √5 true

x > 1 + √5 try x = 5: |5 - 1| < √5 false

so the answer is 1 - √5 < x < 1 + √5

(x-1)^2 < 5

x-1 < 5^0.5

However, the original equation was a square (^2) which is always +ve.

Hence the abs constraint

abs(x-1) < 5^0.5