Solve 13/9x^2+1leqx^2+4/3x

$$\frac{ 13 }{ 9 } { x }^{ 2 } +1 \leq { x }^{ 2 } + \frac{ 4 }{ 3 } x$$

Solve for x

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Subtract x^{2} from both sides.

$$\frac{13}{9}x^{2}+1-x^{2}\leq \frac{4}{3}x$$

Combine \frac{13}{9}x^{2} and -x^{2} to get \frac{4}{9}x^{2}.

$$\frac{4}{9}x^{2}+1\leq \frac{4}{3}x$$

Subtract \frac{4}{3}x from both sides.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x\leq 0$$

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.

$$\frac{4}{9}x^{2}+1-\frac{4}{3}x=0$$

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{4}{9} for a, -\frac{4}{3} for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-\frac{4}{3}\right)±\sqrt{\left(-\frac{4}{3}\right)^{2}-4\times \frac{4}{9}\times 1}}{\frac{4}{9}\times 2}$$

Do the calculations.

$$x=\frac{\frac{4}{3}±0}{\frac{8}{9}}$$

Solutions are the same.

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

Rewrite the inequality by using the obtained solutions.

$$\frac{4}{9}\left(x-\frac{3}{2}\right)^{2}\leq 0$$

Inequality holds for x=\frac{3}{2}.

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