What is (10x^2+4x-1)/2 <0 ?

The solution to (10x^2+4x-1)/2 <0 is -(sqrt(7))/(5sqrt(2))-1/5 <x<(sqrt(7))/(5sqrt(2))-1/5

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(10x^2+4x-1)/2 <0

\frac{10 x ^{2} + 4 x - 1}{2} < 0

- \frac{7}{5 2} - \frac{1}{5} < x < \frac{7}{5 2} - \frac{1}{5}

Interval Notation

(- \frac{7}{5 2} - \frac{1}{5}, \frac{7}{5 2} - \frac{1}{5})

Decimal Notation

- 0.57416 \dots < x < 0.17416 \dots

Solution steps

\frac{10 x ^{2} + 4 x - 1}{2} < 0

Rewrite in standard form

10 x^{2} + 4 x - 1 < 0

Complete the square

10 x^{2} + 4 x - 1 : 10 (x + \frac{1}{5})^{2} - \frac{7}{5}

10 (x + \frac{1}{5})^{2} - \frac{7}{5} < 0

Move

\frac{7}{5}

to the right side

10 (x + \frac{1}{5})^{2} < \frac{7}{5}

Divide both sides by

10

(x + \frac{1}{5})^{2} < \frac{7}{50}

For

u^{n} < a

, if

n

is even then

- \frac{7}{50} < x + \frac{1}{5} < \frac{7}{50}

a < u < b

then

a < u and u < b

- \frac{7}{50} < x + \frac{1}{5} and x + \frac{1}{5} < \frac{7}{50}

- \frac{7}{50} < x + \frac{1}{5} : x > - \frac{7}{5 2} - \frac{1}{5}

x + \frac{1}{5} < \frac{7}{50} : x < \frac{7}{5 2} - \frac{1}{5}

Combine the intervals

x > - \frac{7}{5 2} - \frac{1}{5} and x < \frac{7}{5 2} - \frac{1}{5}

Merge Overlapping Intervals

- \frac{7}{5 2} - \frac{1}{5} < x < \frac{7}{5 2} - \frac{1}{5}

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

2 x^{2} - 4 x + 3 > 0

\frac{( 7 x + 5 )}{( 3 x - 1 )} < 5

x^{2} \cdot (x + 3) \cdot (x - 5)^{3} \cdot (x - 7)^{4} \geq 0

\frac{2 + 3}{( x + 1 )} \geq \frac{2}{x}

What is (10x^2+4x-1)/2 <0 ?
The solution to (10x^2+4x-1)/2 <0 is -(sqrt(7))/(5sqrt(2))-1/5 <x<(sqrt(7))/(5sqrt(2))-1/5