What is 3x^2-11x+2<0 ?

The solution to 3x^2-11x+2<0 is (-sqrt(97)+11)/6 <x<(sqrt(97)+11)/6

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

3x^2-11x+2<0

3 x^{2} - 11 x + 2 < 0

\frac{- 97 + 11}{6} < x < \frac{97 + 11}{6}

Interval Notation

(\frac{- 97 + 11}{6}, \frac{97 + 11}{6})

Decimal Notation

0.19185 \dots < x < 3.47480 \dots

Solution steps

3 x^{2} - 11 x + 2 < 0

Complete the square

3 x^{2} - 11 x + 2 : 3 (x - \frac{11}{6})^{2} - \frac{97}{12}

3 (x - \frac{11}{6})^{2} - \frac{97}{12} < 0

Move

\frac{97}{12}

to the right side

3 (x - \frac{11}{6})^{2} < \frac{97}{12}

Divide both sides by

3

(x - \frac{11}{6})^{2} < \frac{97}{36}

For

u^{n} < a

, if

n

is even then

- \frac{97}{36} < x - \frac{11}{6} < \frac{97}{36}

a < u < b

then

a < u and u < b

- \frac{97}{36} < x - \frac{11}{6} and x - \frac{11}{6} < \frac{97}{36}

- \frac{97}{36} < x - \frac{11}{6} : x > \frac{- 97 + 11}{6}

x - \frac{11}{6} < \frac{97}{36} : x < \frac{97 + 11}{6}

Combine the intervals

x > \frac{- 97 + 11}{6} and x < \frac{97 + 11}{6}

Merge Overlapping Intervals

\frac{- 97 + 11}{6} < x < \frac{97 + 11}{6}

12 - (x^{2} - 101) < - 74 + 6 x^{2}

\frac{( 8 + 4 x )}{( 4 x + x ^{2} )} \leq \frac{2}{x} + \frac{3}{( 4 + x )}

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

(3 x + 1)^{2} + 4 x < 9 x^{2} + x - 4

((x^{2} - 5 x + 6)^{2})^{0} ((x^{2} - 11 x + 24)^{3})^{1} > 0

What is 3x^2-11x+2<0 ?
The solution to 3x^2-11x+2<0 is (-sqrt(97)+11)/6 <x<(sqrt(97)+11)/6