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

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x^2-2x<= 1

Solution

x^{2} - 2 x \leq 1

- 2 + 1 \leq x \leq 2 + 1

Interval Notation

[- 2 + 1, 2 + 1]

Decimal Notation

- 0.41421 \dots \leq x \leq 2.41421 \dots

Solution steps

x^{2} - 2 x \leq 1

Rewrite in standard form

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

Complete the square

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

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

Move

2

to the right side

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

For

u^{n} \leq a

, if

n

is even then

- 2 \leq x - 1 \leq 2

a \leq u \leq b

then

a \leq u and u \leq b

- 2 \leq x - 1 and x - 1 \leq 2

- 2 \leq x - 1 : x \geq - 2 + 1

x - 1 \leq 2 : x \leq 2 + 1

Combine the intervals

x \geq - 2 + 1 and x \leq 2 + 1

Merge Overlapping Intervals

- 2 + 1 \leq x \leq 2 + 1

e^{- 10}

∣ 4 x + 2 ∣ \leq - 8

2 n^{2} - 10 n = 10

4 (3 + \frac{1}{4} c - \frac{1}{2} d)

\frac{10}{5.74} = \frac{x}{2}