The solution to x^2-2x<= 4 is -sqrt(5)+1<= x<= sqrt(5)+1

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x^2-2x<= 4

Solution

x^{2} - 2 x \leq 4

- 5 + 1 \leq x \leq 5 + 1

Interval Notation

[- 5 + 1, 5 + 1]

Decimal Notation

- 1.23606 \dots \leq x \leq 3.23606 \dots

Solution steps

x^{2} - 2 x \leq 4

Rewrite in standard form

x^{2} - 2 x - 4 \leq 0

Complete the square

x^{2} - 2 x - 4 : (x - 1)^{2} - 5

(x - 1)^{2} - 5 \leq 0

Move

5

to the right side

(x - 1)^{2} \leq 5

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x - 1 \leq 5

a \leq u \leq b

then

a \leq u and u \leq b

- 5 \leq x - 1 and x - 1 \leq 5

- 5 \leq x - 1 : x \geq - 5 + 1

x - 1 \leq 5 : x \leq 5 + 1

Combine the intervals

x \geq - 5 + 1 and x \leq 5 + 1

Merge Overlapping Intervals

- 5 + 1 \leq x \leq 5 + 1

2 x^{2} + 4 x + 8 = 0

\frac{9 - i}{- 4 + 5 i}

\frac{( x ^{2} + y ^{2} )}{x ^{2} - y ^{2}} + \frac{y}{x + y} - \frac{x}{x - y}

\frac{n ^{2} + 2 n - 24}{n ^{2} - 11 n + 28}

- 9 x^{2} - 6 x - 1 = 0