Solve 4p^2-8p+3<0

$$4 p ^ { 2 } - 8 p + 3 < 0$$

Solve for p

$$p\in \left(\frac{1}{2},\frac{3}{2}\right)$$

Steps Using the Quadratic Formula

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$4p^{2}-8p+3=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, -8 for b, and 3 for c in the quadratic formula.

$$p=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 4\times 3}}{2\times 4}$$

Do the calculations.

$$p=\frac{8±4}{8}$$

Solve the equation p=\frac{8±4}{8} when ± is plus and when ± is minus.

$$p=\frac{3}{2} p=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$4\left(p-\frac{3}{2}\right)\left(p-\frac{1}{2}\right)<0$$

For the product to be negative, p-\frac{3}{2} and p-\frac{1}{2} have to be of the opposite signs. Consider the case when p-\frac{3}{2} is positive and p-\frac{1}{2} is negative.

$$p-\frac{3}{2}>0 p-\frac{1}{2}<0$$

This is false for any p.

$$p\in \emptyset$$

Consider the case when p-\frac{1}{2} is positive and p-\frac{3}{2} is negative.

$$p-\frac{1}{2}>0 p-\frac{3}{2}<0$$

The solution satisfying both inequalities is p\in \left(\frac{1}{2},\frac{3}{2}\right).

$$p\in \left(\frac{1}{2},\frac{3}{2}\right)$$

The final solution is the union of the obtained solutions.

$$p\in \left(\frac{1}{2},\frac{3}{2}\right)$$