What is 2+2x+2x^2<= 6+x+x^2 ?

The solution to 2+2x+2x^2<= 6+x+x^2 is (-sqrt(17)-1)/2 <= x<= (sqrt(17)-1)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2+2x+2x^2<= 6+x+x^2

2 + 2 x + 2 x^{2} \leq 6 + x + x^{2}

\frac{- 17 - 1}{2} \leq x \leq \frac{17 - 1}{2}

Interval Notation

[\frac{- 17 - 1}{2}, \frac{17 - 1}{2}]

Decimal Notation

- 2.56155 \dots \leq x \leq 1.56155 \dots

Solution steps

2 + 2 x + 2 x^{2} \leq 6 + x + x^{2}

Rewrite in standard form

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

Complete the square

x^{2} + x - 4 : (x + \frac{1}{2})^{2} - \frac{17}{4}

(x + \frac{1}{2})^{2} - \frac{17}{4} \leq 0

Move

\frac{17}{4}

to the right side

(x + \frac{1}{2})^{2} \leq \frac{17}{4}

For

u^{n} \leq a

, if

n

is even then

- \frac{17}{4} \leq x + \frac{1}{2} \leq \frac{17}{4}

a \leq u \leq b

then

a \leq u and u \leq b

- \frac{17}{4} \leq x + \frac{1}{2} and x + \frac{1}{2} \leq \frac{17}{4}

- \frac{17}{4} \leq x + \frac{1}{2} : x \geq \frac{- 17 - 1}{2}

x + \frac{1}{2} \leq \frac{17}{4} : x \leq \frac{17 - 1}{2}

Combine the intervals

x \geq \frac{- 17 - 1}{2} and x \leq \frac{17 - 1}{2}

Merge Overlapping Intervals

\frac{- 17 - 1}{2} \leq x \leq \frac{17 - 1}{2}

\frac{∣ 3 + x ∣}{7} \leq 5

x^{3} + 8 x^{2} + 5 x - 14

\frac{1}{2} 2

x^{2} - 4 x - 3 = 0

- 3 - 1

What is 2+2x+2x^2<= 6+x+x^2 ?
The solution to 2+2x+2x^2<= 6+x+x^2 is (-sqrt(17)-1)/2 <= x<= (sqrt(17)-1)/2