The solution to x^2+2x-1<0 is -sqrt(2)-1<x<sqrt(2)-1

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x^2+2x-1<0

Solution

x^{2} + 2 x - 1 < 0

- 2 - 1 < x < 2 - 1

Solution steps

x^{2} + 2 x - 1 < 0

Complete the square

x^{2} + 2 x - 1 : (x + 1)^{2} - 2

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

Move

2

to the right side

(x + 1)^{2} < 2

For

u^{n} < a

, if

n

is even then

- 2 < x + 1 < 2

a < u < b

then

a < u and u < b

- 2 < x + 1 and x + 1 < 2

- 2 < x + 1 : x > - 2 - 1

x + 1 < 2 : x < 2 - 1

Combine the intervals

x > - 2 - 1 and x < 2 - 1

Merge Overlapping Intervals

- 2 - 1 < x < 2 - 1

3 \cdot 7

[(2 d + 1)^{(\frac{1}{2})}] + [(d)^{(\frac{1}{2})}] \leq 5

7 x - 5 = 23

4 = 4 (x + 10)

\frac{5 + e}{2} = - 10