Solve (2/3x-4)^2geqx^2

$$( \frac { 2 } { 3 } x - 4 ) ^ { 2 } \geq x ^ { 2 }$$

Solve for x

$$x\in \begin{bmatrix}-12,\frac{12}{5}\end{bmatrix}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\frac{2}{3}x-4\right)^{2}.

$$\frac{4}{9}x^{2}-\frac{16}{3}x+16\geq x^{2}$$

Subtract x^{2} from both sides.

$$\frac{4}{9}x^{2}-\frac{16}{3}x+16-x^{2}\geq 0$$

Combine \frac{4}{9}x^{2} and -x^{2} to get -\frac{5}{9}x^{2}.

$$-\frac{5}{9}x^{2}-\frac{16}{3}x+16\geq 0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -\frac{5}{9}x^{2}-\frac{16}{3}x+16 positive. Since -1 is negative, the inequality direction is changed.

$$\frac{5}{9}x^{2}+\frac{16}{3}x-16\leq 0$$

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.

$$\frac{5}{9}x^{2}+\frac{16}{3}x-16=0$$

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 \frac{5}{9} for a, \frac{16}{3} for b, and -16 for c in the quadratic formula.

$$x=\frac{-\frac{16}{3}±\sqrt{\left(\frac{16}{3}\right)^{2}-4\times \frac{5}{9}\left(-16\right)}}{\frac{5}{9}\times 2}$$

Do the calculations.

$$x=\frac{-\frac{16}{3}±8}{\frac{10}{9}}$$

Solve the equation x=\frac{-\frac{16}{3}±8}{\frac{10}{9}} when ± is plus and when ± is minus.

$$x=\frac{12}{5} x=-12$$

Rewrite the inequality by using the obtained solutions.

$$\frac{5}{9}\left(x-\frac{12}{5}\right)\left(x+12\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{12}{5} and x+12 has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{12}{5}\geq 0 and x+12\leq 0.

$$x-\frac{12}{5}\geq 0 x+12\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{12}{5}\leq 0 and x+12\geq 0.

$$x+12\geq 0 x-\frac{12}{5}\leq 0$$

The solution satisfying both inequalities is x\in \left[-12,\frac{12}{5}\right].

$$x\in \begin{bmatrix}-12,\frac{12}{5}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}-12,\frac{12}{5}\end{bmatrix}$$