The solution to 8-3x-x^2>0 is (-sqrt(41)-3)/2 <x<(sqrt(41)-3)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

8-3x-x^2>0

Solution

8 - 3 x - x^{2} > 0

\frac{- 41 - 3}{2} < x < \frac{41 - 3}{2}

Interval Notation

(\frac{- 41 - 3}{2}, \frac{41 - 3}{2})

Decimal Notation

- 4.70156 \dots < x < 1.70156 \dots

Solution steps

8 - 3 x - x^{2} > 0

Complete the square

8 - 3 x - x^{2} : - (x + \frac{3}{2})^{2} + \frac{41}{4}

- (x + \frac{3}{2})^{2} + \frac{41}{4} > 0

Move

\frac{41}{4}

to the right side

- (x + \frac{3}{2})^{2} > - \frac{41}{4}

Multiply both sides by

- 1

(x + \frac{3}{2})^{2} < \frac{41}{4}

For

u^{n} < a

, if

n

is even then

- \frac{41}{4} < x + \frac{3}{2} < \frac{41}{4}

a < u < b

then

a < u and u < b

- \frac{41}{4} < x + \frac{3}{2} and x + \frac{3}{2} < \frac{41}{4}

- \frac{41}{4} < x + \frac{3}{2} : x > \frac{- 41 - 3}{2}

x + \frac{3}{2} < \frac{41}{4} : x < \frac{41 - 3}{2}

Combine the intervals

x > \frac{- 41 - 3}{2} and x < \frac{41 - 3}{2}

Merge Overlapping Intervals

\frac{- 41 - 3}{2} < x < \frac{41 - 3}{2}

\frac{3}{6} + \frac{4}{6}

- 3 x - 21 \geq 10

16 x^{3} + 8 x^{2} + x

\frac{1}{4} \cdot \frac{2}{3}

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

What is 8-3x-x^2>0 ?
The solution to 8-3x-x^2>0 is (-sqrt(41)-3)/2 <x<(sqrt(41)-3)/2