The solution to x^2-2x<= 2 is -sqrt(3)+1<= x<= sqrt(3)+1

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x^2-2x<= 2

Solution

x^{2} - 2 x \leq 2

- 3 + 1 \leq x \leq 3 + 1

Interval Notation

[- 3 + 1, 3 + 1]

Decimal Notation

- 0.73205 \dots \leq x \leq 2.73205 \dots

Solution steps

x^{2} - 2 x \leq 2

Rewrite in standard form

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

Complete the square

x^{2} - 2 x - 2 : (x - 1)^{2} - 3

(x - 1)^{2} - 3 \leq 0

Move

3

to the right side

(x - 1)^{2} \leq 3

For

u^{n} \leq a

, if

n

is even then

- 3 \leq x - 1 \leq 3

a \leq u \leq b

then

a \leq u and u \leq b

- 3 \leq x - 1 and x - 1 \leq 3

- 3 \leq x - 1 : x \geq - 3 + 1

x - 1 \leq 3 : x \leq 3 + 1

Combine the intervals

x \geq - 3 + 1 and x \leq 3 + 1

Merge Overlapping Intervals

- 3 + 1 \leq x \leq 3 + 1

3 x^{2} - x - 3 = 0

\frac{3}{4} \cdot 16

∣ x - 7 ∣ = - 1

(9 - 2 x)^{2} \leq (4 x)^{2}

x (b + c) + 3 b + 3 c