What is -x^2+2x>(4x-1)/4 ?

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

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-x^2+2x>(4x-1)/4

- x^{2} + 2 x > \frac{4 x - 1}{4}

- \frac{1}{2} + \frac{1}{2} < x < \frac{1}{2} + \frac{1}{2}

Interval Notation

(- \frac{1}{2} + \frac{1}{2}, \frac{1}{2} + \frac{1}{2})

Decimal Notation

- 0.20710 \dots < x < 1.20710 \dots

Solution steps

- x^{2} + 2 x > \frac{4 x - 1}{4}

Rewrite in standard form

- 4 x^{2} + 4 x + 1 > 0

Complete the square

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

- 4 (x - \frac{1}{2})^{2} + 2 > 0

Move

2

to the right side

- 4 (x - \frac{1}{2})^{2} > - 2

Multiply both sides by

- 1

4 (x - \frac{1}{2})^{2} < 2

Divide both sides by

4

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

For

u^{n} < a

, if

n

is even then

- \frac{1}{2} < x - \frac{1}{2} < \frac{1}{2}

a < u < b

then

a < u and u < b

- \frac{1}{2} < x - \frac{1}{2} and x - \frac{1}{2} < \frac{1}{2}

- \frac{1}{2} < x - \frac{1}{2} : x > - \frac{1}{2} + \frac{1}{2}

x - \frac{1}{2} < \frac{1}{2} : x < \frac{1}{2} + \frac{1}{2}

Combine the intervals

x > - \frac{1}{2} + \frac{1}{2} and x < \frac{1}{2} + \frac{1}{2}

Merge Overlapping Intervals

- \frac{1}{2} + \frac{1}{2} < x < \frac{1}{2} + \frac{1}{2}

(2 x - 1) (x^{2} - x - 6) < 0

x^{3} - 48 \leq 4 x (3 x - 11)

(x + 2)^{2} > 2

\frac{2 x - 1}{x + 1} < 0

\frac{( 3 n ^{2} )}{( n ^{2} - 1 )} < 0

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