The solution to x^2-3x+1<0 is (-sqrt(5)+3)/2 <x<(sqrt(5)+3)/2

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x^2-3x+1<0

Solution

x^{2} - 3 x + 1 < 0

\frac{- 5 + 3}{2} < x < \frac{5 + 3}{2}

Interval Notation

(\frac{- 5 + 3}{2}, \frac{5 + 3}{2})

Decimal Notation

0.38196 \dots < x < 2.61803 \dots

Solution steps

x^{2} - 3 x + 1 < 0

Complete the square

x^{2} - 3 x + 1 : (x - \frac{3}{2})^{2} - \frac{5}{4}

(x - \frac{3}{2})^{2} - \frac{5}{4} < 0

Move

\frac{5}{4}

to the right side

(x - \frac{3}{2})^{2} < \frac{5}{4}

For

u^{n} < a

, if

n

is even then

- \frac{5}{4} < x - \frac{3}{2} < \frac{5}{4}

a < u < b

then

a < u and u < b

- \frac{5}{4} < x - \frac{3}{2} and x - \frac{3}{2} < \frac{5}{4}

- \frac{5}{4} < x - \frac{3}{2} : x > \frac{- 5 + 3}{2}

x - \frac{3}{2} < \frac{5}{4} : x < \frac{5 + 3}{2}

Combine the intervals

x > \frac{- 5 + 3}{2} and x < \frac{5 + 3}{2}

Merge Overlapping Intervals

\frac{- 5 + 3}{2} < x < \frac{5 + 3}{2}

4 k (k - 1) > 0

(4 (a - 2))^{2} - 4 (a^{2}) > 0

\frac{x + 1}{2} > \frac{1}{x + 2}

(n - 3)^{2} - (n + 4) (n - 4) < 43

12 x^{2} + x - 6 < 0

What is x^2-3x+1<0 ?
The solution to x^2-3x+1<0 is (-sqrt(5)+3)/2 <x<(sqrt(5)+3)/2