Solve 1/2x(2x-3/2)>0

$$\frac { 1 } { 2 } x ( 2 x - \frac { 3 } { 2 } ) > 0$$

Solve for x

$$x\in \left(-\infty,0\right)\cup \left(\frac{3}{4},\infty\right)$$

Solution Steps

Use the distributive property to multiply \frac{1}{2}x by 2x-\frac{3}{2}.

$$\frac{1}{2}x\times 2x+\frac{1}{2}x\left(-\frac{3}{2}\right)>0$$

Multiply x and x to get x^{2}.

$$\frac{1}{2}x^{2}\times 2+\frac{1}{2}x\left(-\frac{3}{2}\right)>0$$

Cancel out 2 and 2.

$$x^{2}+\frac{1}{2}x\left(-\frac{3}{2}\right)>0$$

Multiply \frac{1}{2} times -\frac{3}{2} by multiplying numerator times numerator and denominator times denominator.

$$x^{2}+\frac{1\left(-3\right)}{2\times 2}x>0$$

Do the multiplications in the fraction \frac{1\left(-3\right)}{2\times 2}.

$$x^{2}+\frac{-3}{4}x>0$$

Fraction \frac{-3}{4} can be rewritten as -\frac{3}{4} by extracting the negative sign.

$$x^{2}-\frac{3}{4}x>0$$

Factor out x.

$$\frac{1}{4}x\left(4x-3\right)>0$$

For the product to be positive, x and x-\frac{3}{4} have to be both negative or both positive. Consider the case when x and x-\frac{3}{4} are both negative.

$$x<0 x-\frac{3}{4}<0$$

The solution satisfying both inequalities is x<0.

$$x<0$$

Consider the case when x and x-\frac{3}{4} are both positive.

$$x-\frac{3}{4}>0 x>0$$

The solution satisfying both inequalities is x>\frac{3}{4}.

$$x>\frac{3}{4}$$

The final solution is the union of the obtained solutions.

$$x<0\text{; }x>\frac{3}{4}$$