What is x^2-2x-4<= 0 ?

The solution to x^2-2x-4<= 0 is -sqrt(5)+1<= x<= sqrt(5)+1

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x^2-2x-4<= 0

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

- 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 - 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

\frac{- 2}{x} < 0

(3 m - m^{2})^{2} > 0

\frac{4}{( x + 2 )} \geq 3 - x

(2 x - 6) (- 2 x + 8) > 0

\frac{( ( x - 5 ) ( x - 9 ) )}{( ( x - 3 ) ( x - 7 ) )} \leq 0

What is x^2-2x-4<= 0 ?
The solution to x^2-2x-4<= 0 is -sqrt(5)+1<= x<= sqrt(5)+1