What is 2x^2-4x-1<4x ?

The solution to 2x^2-4x-1<4x is -3/(sqrt(2))+2<x< 3/(sqrt(2))+2

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2x^2-4x-1<4x

2 x^{2} - 4 x - 1 < 4 x

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

Interval Notation

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

Decimal Notation

- 0.12132 \dots < x < 4.12132 \dots

Solution steps

2 x^{2} - 4 x - 1 < 4 x

Rewrite in standard form

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

Complete the square

2 x^{2} - 8 x - 1 : 2 (x - 2)^{2} - 9

2 (x - 2)^{2} - 9 < 0

Move

9

to the right side

2 (x - 2)^{2} < 9

Divide both sides by

2

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

For

u^{n} < a

, if

n

is even then

- \frac{9}{2} < x - 2 < \frac{9}{2}

a < u < b

then

a < u and u < b

- \frac{9}{2} < x - 2 and x - 2 < \frac{9}{2}

- \frac{9}{2} < x - 2 : x > - \frac{3}{2} + 2

x - 2 < \frac{9}{2} : x < \frac{3}{2} + 2

Combine the intervals

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

Merge Overlapping Intervals

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

- 2 (y - 4) = 20 - 2 y - 12

- 6 x + 35 = - 6 x - 35

(x - 3) (x - 5) = 0

5 a^{2} b - 10 a^{2} b^{2}

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

What is 2x^2-4x-1<4x ?
The solution to 2x^2-4x-1<4x is -3/(sqrt(2))+2<x< 3/(sqrt(2))+2