What is x+sqrt(x^2-1)<5 ?

The solution to x+sqrt(x^2-1)<5 is x<=-1\lor 1<= x< 13/5

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x+sqrt(x^2-1)<5

x + x^{2} - 1 < 5

x \leq - 1 or 1 \leq x < \frac{13}{5}

Interval Notation

(- \infty, - 1] \cup [1, \frac{13}{5})

Decimal Notation

x \leq - 1 or 1 \leq x < 2.6

Solution steps

x + x^{2} - 1 < 5

Subtract

x

from both sides

x + x^{2} - 1 - x < 5 - x

Simplify

x^{2} - 1 < 5 - x

f (x) < g (x)

then

g (x) > 0

5 - x > 0 : x < 5

x^{2} - 1 < 5 - x : x < \frac{13}{5}

Find singularity points

Find non-negative values for radicals:

x \leq - 1 or x \geq 1

Combine the intervals

x < 5 and x < \frac{13}{5} and (x \leq - 1 or x \geq 1)

Merge Overlapping Intervals

x \leq - 1 or 1 \leq x < \frac{13}{5}

Verify Solutions:

x \leq - 1

True

, 1 \leq x < \frac{13}{5}

True

The solution is

x \leq - 1 or 1 \leq x < \frac{13}{5}

124

y^{2} + 9 y + 20

- 15 x + 4 \leq 109 or - 6 x + 70 > - 2

(1400000000) \cdot (\frac{100 - 2.2}{100})^{32}

2 (7 x - 1) - 3 (3 x - 6) - 5 (11 x + 6) = 196

What is x+sqrt(x^2-1)<5 ?
The solution to x+sqrt(x^2-1)<5 is x<=-1\lor 1<= x< 13/5