What is (5x^2-3)/4+x/2 >3x-1 ?

The solution to (5x^2-3)/4+x/2 >3x-1 is x 2/(sqrt(5))+1

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(5x^2-3)/4+x/2 >3x-1

\frac{5 x ^{2} - 3}{4} + \frac{x}{2} > 3 x - 1

x < - \frac{2}{5} + 1 or x > \frac{2}{5} + 1

Interval Notation

(- \infty, - \frac{2}{5} + 1) \cup (\frac{2}{5} + 1, \infty)

Decimal Notation

x < 0.10557 \dots or x > 1.89442 \dots

Solution steps

\frac{5 x ^{2} - 3}{4} + \frac{x}{2} > 3 x - 1

Rewrite in standard form

5 x^{2} - 10 x + 1 > 0

Complete the square

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

5 (x - 1)^{2} - 4 > 0

Move

4

to the right side

5 (x - 1)^{2} > 4

Divide both sides by

5

(x - 1)^{2} > \frac{4}{5}

For

u^{n} > a

, if

n

is even then

x - 1 < - \frac{4}{5} or x - 1 > \frac{4}{5}

x - 1 < - \frac{4}{5} : x < - \frac{2}{5} + 1

x - 1 > \frac{4}{5} : x > \frac{2}{5} + 1

Combine the intervals

x < - \frac{2}{5} + 1 or x > \frac{2}{5} + 1

7 - r = 8 - r - r

x^{5} + 7 x^{3}

∣ 3 x - 1 ∣ \geq 4

- 1 < x + 3 < 5

12 x^{2} + 48 x - 826 = 0

What is (5x^2-3)/4+x/2 >3x-1 ?
The solution to (5x^2-3)/4+x/2 >3x-1 is x<-2/(sqrt(5))+1\lor x> 2/(sqrt(5))+1