What is 4x-x^2+8>= 0 ?

The solution to 4x-x^2+8>= 0 is -2sqrt(3)+2<= x<= 2sqrt(3)+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4x-x^2+8>= 0

4 x - x^{2} + 8 \geq 0

- 23 + 2 \leq x \leq 23 + 2

Interval Notation

[- 23 + 2, 23 + 2]

Decimal Notation

- 1.46410 \dots \leq x \leq 5.46410 \dots

Solution steps

4 x - x^{2} + 8 \geq 0

Complete the square

4 x - x^{2} + 8 : - (x - 2)^{2} + 12

- (x - 2)^{2} + 12 \geq 0

Move

12

to the right side

- (x - 2)^{2} \geq - 12

Multiply both sides by

- 1

(x - 2)^{2} \leq 12

For

u^{n} \leq a

, if

n

is even then

- 12 \leq x - 2 \leq 12

a \leq u \leq b

then

a \leq u and u \leq b

- 12 \leq x - 2 and x - 2 \leq 12

- 12 \leq x - 2 : x \geq - 23 + 2

x - 2 \leq 12 : x \leq 23 + 2

Combine the intervals

x \geq - 23 + 2 and x \leq 23 + 2

Merge Overlapping Intervals

- 23 + 2 \leq x \leq 23 + 2

\frac{x - 3}{2 x - 4} > 0

(x + 4)^{2} < 0

t^{4} + t^{3} - 12 t^{2} < 0

\frac{3}{( 2 - x )} \leq \frac{1}{( x + 4 )}

\frac{( 2 x - 1 )}{( x + 1 )} \geq 6

What is 4x-x^2+8>= 0 ?
The solution to 4x-x^2+8>= 0 is -2sqrt(3)+2<= x<= 2sqrt(3)+2