Solve (x-1/4)^2leq25/16

$$( x - \frac { 1 } { 4 } ) ^ { 2 } \leq \frac { 25 } { 16 }$$

Solve for x

$$x\in \begin{bmatrix}-1,\frac{3}{2}\end{bmatrix}$$

Solution Steps

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.

$$\left(x-\frac{1}{4}\right)^{2}=0$$

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 1 for a, -\frac{1}{2} for b, and -\frac{3}{2} for c in the quadratic formula.

$$x=\frac{-\left(-\frac{1}{2}\right)±\sqrt{\left(-\frac{1}{2}\right)^{2}-4\times 1\left(-\frac{3}{2}\right)}}{2}$$

Do the calculations.

$$x=\frac{\frac{1}{2}±\frac{5}{2}}{2}$$

Solve the equation x=\frac{\frac{1}{2}±\frac{5}{2}}{2} when ± is plus and when ± is minus.

$$x=\frac{3}{2} x=-1$$

Rewrite the inequality by using the obtained solutions.

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

For the product to be ≤0, one of the values x-\frac{3}{2} and x+1 has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{3}{2}\geq 0 and x+1\leq 0.

$$x-\frac{3}{2}\geq 0 x+1\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{3}{2}\leq 0 and x+1\geq 0.

$$x+1\geq 0 x-\frac{3}{2}\leq 0$$

The solution satisfying both inequalities is x\in \left[-1,\frac{3}{2}\right].

$$x\in \begin{bmatrix}-1,\frac{3}{2}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-1,\frac{3}{2}\end{bmatrix}$$