What is 2x^2-5x+1<0 ?

The solution to 2x^2-5x+1<0 is (-sqrt(17)+5)/4 <x<(sqrt(17)+5)/4

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2x^2-5x+1<0

2 x^{2} - 5 x + 1 < 0

\frac{- 17 + 5}{4} < x < \frac{17 + 5}{4}

Interval Notation

(\frac{- 17 + 5}{4}, \frac{17 + 5}{4})

Decimal Notation

0.21922 \dots < x < 2.28077 \dots

Solution steps

2 x^{2} - 5 x + 1 < 0

Complete the square

2 x^{2} - 5 x + 1 : 2 (x - \frac{5}{4})^{2} - \frac{17}{8}

2 (x - \frac{5}{4})^{2} - \frac{17}{8} < 0

Move

\frac{17}{8}

to the right side

2 (x - \frac{5}{4})^{2} < \frac{17}{8}

Divide both sides by

2

(x - \frac{5}{4})^{2} < \frac{17}{16}

For

u^{n} < a

, if

n

is even then

- \frac{17}{16} < x - \frac{5}{4} < \frac{17}{16}

a < u < b

then

a < u and u < b

- \frac{17}{16} < x - \frac{5}{4} and x - \frac{5}{4} < \frac{17}{16}

- \frac{17}{16} < x - \frac{5}{4} : x > \frac{- 17 + 5}{4}

x - \frac{5}{4} < \frac{17}{16} : x < \frac{17 + 5}{4}

Combine the intervals

x > \frac{- 17 + 5}{4} and x < \frac{17 + 5}{4}

Merge Overlapping Intervals

\frac{- 17 + 5}{4} < x < \frac{17 + 5}{4}

21 x + 5 x^{2} = - 4

- - 13

0 = 3 t^{2} - 36 t + 96

9 a^{5} b - 12 a^{2} b^{3} + 15 a b^{2} - 18 a^{3} b^{4}

(\frac{1}{27})^{2 x + 1} \leq (\frac{1}{243})^{3 x - 2}

What is 2x^2-5x+1<0 ?
The solution to 2x^2-5x+1<0 is (-sqrt(17)+5)/4 <x<(sqrt(17)+5)/4