What is -k^2+2k+1<= 0 ?

The solution to -k^2+2k+1 = sqrt(2)+1

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-k^2+2k+1<= 0

- k^{2} + 2 k + 1 \leq 0

k \leq - 2 + 1 or k \geq 2 + 1

Interval Notation

(- \infty, - 2 + 1] \cup [2 + 1, \infty)

Decimal Notation

k \leq - 0.41421 \dots or k \geq 2.41421 \dots

Solution steps

- k^{2} + 2 k + 1 \leq 0

Complete the square

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

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

Move

2

to the right side

- (k - 1)^{2} \leq - 2

Multiply both sides by

- 1

(k - 1)^{2} \geq 2

For

u^{n} \geq a

, if

n

is even then

k - 1 \leq - 2 or k - 1 \geq 2

k - 1 \leq - 2 : k \leq - 2 + 1

k - 1 \geq 2 : k \geq 2 + 1

Combine the intervals

k \leq - 2 + 1 or k \geq 2 + 1

7 x^{2} - 31 x - 20

(5^{3})^{\frac{1}{8}}

- 17 + \frac{q}{8} = 13

(- 9)^{2}

2.8 x + 4 \leq - x - 2.1 (- 2 x - 4)

What is -k^2+2k+1<= 0 ?
The solution to -k^2+2k+1<= 0 is k<=-sqrt(2)+1\lor k>= sqrt(2)+1