What is sqrt(x+1)sqrt(x-1)<= 0 ?

The solution to sqrt(x+1)sqrt(x-1)<= 0 is x=1

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sqrt(x+1)sqrt(x-1)<= 0

x + 1 x - 1 \leq 0

x = 1

Interval Notation

x = 1

Solution steps

x + 1 x - 1 \leq 0

Apply radical rule:

a b = a \cdot b,

assuming

a \geq 0, b \geq 0

x + 1 x - 1 = (x + 1) (x - 1)

(x + 1) (x - 1) \leq 0

Square both sides

((x + 1) (x - 1))^{2} \leq 0^{2}

Simplify

x^{2} - 1 \leq 0

x^{2} - 1 \leq 0 : - 1 \leq x \leq 1

- 1 \leq x \leq 1

Find singularity points

Find non-negative values for radicals:

x \geq 1

Combine the intervals

- 1 \leq x \leq 1 and x \geq 1

Merge Overlapping Intervals

x = 1

x^{2} - \frac{2}{5} x + \frac{1}{25}

(- 512)^{\frac{1}{3}}

\frac{m ^{2} - 6 m + 8}{2 m - 2} \cdot \frac{10}{m - 4}

(x + 3)^{2} (x - 4)^{3} (2 x + 1)^{2} < 0

∣ 8 x - 2 ∣ \geq 2