What is x^2-8x<=-11 ?

The solution to x^2-8x<=-11 is -sqrt(5)+4<= x<= sqrt(5)+4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

x^2-8x<=-11

x^{2} - 8 x \leq - 11

- 5 + 4 \leq x \leq 5 + 4

Interval Notation

[- 5 + 4, 5 + 4]

Decimal Notation

1.76393 \dots \leq x \leq 6.23606 \dots

Solution steps

x^{2} - 8 x \leq - 11

Rewrite in standard form

x^{2} - 8 x + 11 \leq 0

Complete the square

x^{2} - 8 x + 11 : (x - 4)^{2} - 5

(x - 4)^{2} - 5 \leq 0

Move

5

to the right side

(x - 4)^{2} \leq 5

For

u^{n} \leq a

, if

n

is even then

- 5 \leq x - 4 \leq 5

a \leq u \leq b

then

a \leq u and u \leq b

- 5 \leq x - 4 and x - 4 \leq 5

- 5 \leq x - 4 : x \geq - 5 + 4

x - 4 \leq 5 : x \leq 5 + 4

Combine the intervals

x \geq - 5 + 4 and x \leq 5 + 4

Merge Overlapping Intervals

- 5 + 4 \leq x \leq 5 + 4

- 2 (\frac{1}{2})^{- 2}

\frac{3}{2} \cdot \frac{2}{3}

(\frac{3}{8})^{- 1}

3 (x - 4) - 7 (x + 2) = - 2 (x + 18)

\frac{59 n}{99} \cdot \frac{80}{33 n}