Solve left(2-3xright)^2geq1

$${ \left(2-3x \right) }^{ 2 } \geq 1$$

Solve for x

$$x\in (-\infty,\frac{1}{3}]\cup [1,\infty)$$

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.

$$\left(2-3x\right)^{2}=0$$

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 9 for a, -12 for b, and 3 for c in the quadratic formula.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 9\times 3}}{2\times 9}$$

Do the calculations.

$$x=\frac{12±6}{18}$$

Solve the equation x=\frac{12±6}{18} when ± is plus and when ± is minus.

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

Rewrite the inequality by using the obtained solutions.

$$9\left(x-1\right)\left(x-\frac{1}{3}\right)\geq 0$$

For the product to be ≥0, x-1 and x-\frac{1}{3} have to be both ≤0 or both ≥0. Consider the case when x-1 and x-\frac{1}{3} are both ≤0.

$$x-1\leq 0 x-\frac{1}{3}\leq 0$$

The solution satisfying both inequalities is x\leq \frac{1}{3}.

$$x\leq \frac{1}{3}$$

Consider the case when x-1 and x-\frac{1}{3} are both ≥0.

$$x-\frac{1}{3}\geq 0 x-1\geq 0$$

The solution satisfying both inequalities is x\geq 1.

$$x\geq 1$$

The final solution is the union of the obtained solutions.

$$x\leq \frac{1}{3}\text{; }x\geq 1$$