What is 3x^2-8x+2<0 ?

The solution to 3x^2-8x+2<0 is (-sqrt(10)+4)/3 <x<(sqrt(10)+4)/3

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3x^2-8x+2<0

3 x^{2} - 8 x + 2 < 0

\frac{- 10 + 4}{3} < x < \frac{10 + 4}{3}

Interval Notation

(\frac{- 10 + 4}{3}, \frac{10 + 4}{3})

Decimal Notation

0.27924 \dots < x < 2.38742 \dots

Solution steps

3 x^{2} - 8 x + 2 < 0

Complete the square

3 x^{2} - 8 x + 2 : 3 (x - \frac{4}{3})^{2} - \frac{10}{3}

3 (x - \frac{4}{3})^{2} - \frac{10}{3} < 0

Move

\frac{10}{3}

to the right side

3 (x - \frac{4}{3})^{2} < \frac{10}{3}

Divide both sides by

3

(x - \frac{4}{3})^{2} < \frac{10}{9}

For

u^{n} < a

, if

n

is even then

- \frac{10}{9} < x - \frac{4}{3} < \frac{10}{9}

a < u < b

then

a < u and u < b

- \frac{10}{9} < x - \frac{4}{3} and x - \frac{4}{3} < \frac{10}{9}

- \frac{10}{9} < x - \frac{4}{3} : x > \frac{- 10 + 4}{3}

x - \frac{4}{3} < \frac{10}{9} : x < \frac{10 + 4}{3}

Combine the intervals

x > \frac{- 10 + 4}{3} and x < \frac{10 + 4}{3}

Merge Overlapping Intervals

\frac{- 10 + 4}{3} < x < \frac{10 + 4}{3}

x (x - 1) (x - 2) (x - 3) > 0

x^{5} - 7 x^{4} + 13 x^{3} - x^{2} - 14 x + 8 \leq 0

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

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

\frac{( x ^{2} + 6 )}{( 5 x )} \geq 1

What is 3x^2-8x+2<0 ?
The solution to 3x^2-8x+2<0 is (-sqrt(10)+4)/3 <x<(sqrt(10)+4)/3