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

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x^2<= 2x+1

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

Solution

x^{2} \leq 2 x + 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} \leq 2 x + 1

Rewrite in standard form

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

Complete the square

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

\frac{100}{\frac{2}{2}} = \frac{x}{\frac{3}{2}}

49 x^{2} - 42 x + 9 = 0

\frac{e}{- 2} > - 3

\frac{4}{n - 4} - \frac{2 - n}{n}

64 \cdot \frac{1}{4}