The solution to x^2-8x+1<0 is -sqrt(15)+4<x<sqrt(15)+4

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x^2-8x+1<0

Solution

x^{2} - 8 x + 1 < 0

- 15 + 4 < x < 15 + 4

Interval Notation

(- 15 + 4, 15 + 4)

Decimal Notation

0.12701 \dots < x < 7.87298 \dots

Solution steps

x^{2} - 8 x + 1 < 0

Complete the square

x^{2} - 8 x + 1 : (x - 4)^{2} - 15

(x - 4)^{2} - 15 < 0

Move

15

to the right side

(x - 4)^{2} < 15

For

u^{n} < a

, if

n

is even then

- 15 < x - 4 < 15

a < u < b

then

a < u and u < b

- 15 < x - 4 and x - 4 < 15

- 15 < x - 4 : x > - 15 + 4

x - 4 < 15 : x < 15 + 4

Combine the intervals

x > - 15 + 4 and x < 15 + 4

Merge Overlapping Intervals

- 15 + 4 < x < 15 + 4

\frac{5}{( 3 x + 1 )} - \frac{20}{( 9 x ^{2} - 1 )} < \frac{2}{( 3 x - 1 )}

\frac{x}{( x - 5 )} \geq \frac{1}{2}

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

9 x^{2} - 8 x - 1 > 0

\frac{1}{3 ( 9 x + 27 )} > x + 33