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

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x^2-5x+2<0

Solution

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

\frac{- 17 + 5}{2} < x < \frac{17 + 5}{2}

Interval Notation

(\frac{- 17 + 5}{2}, \frac{17 + 5}{2})

Decimal Notation

0.43844 \dots < x < 4.56155 \dots

Solution steps

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

Complete the square

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

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

Move

\frac{17}{4}

to the right side

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

For

u^{n} < a

, if

n

is even then

- \frac{17}{4} < x - \frac{5}{2} < \frac{17}{4}

a < u < b

then

a < u and u < b

- \frac{17}{4} < x - \frac{5}{2} and x - \frac{5}{2} < \frac{17}{4}

- \frac{17}{4} < x - \frac{5}{2} : x > \frac{- 17 + 5}{2}

x - \frac{5}{2} < \frac{17}{4} : x < \frac{17 + 5}{2}

Combine the intervals

x > \frac{- 17 + 5}{2} and x < \frac{17 + 5}{2}

Merge Overlapping Intervals

\frac{- 17 + 5}{2} < x < \frac{17 + 5}{2}

\frac{2}{3} (x + 5) < \frac{5}{6} (2 x - 8)

- 10 \cdot 0.02

3 + \frac{1}{4}

x^{2} - 7 x + 7 = 0

\frac{8 !}{2 !}

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