Solve 3x^2-7x+1leq0

$$3 x ^ { 2 } - 7 x + 1 \leq 0$$

Solve for x

$$x\in \begin{bmatrix}\frac{7-\sqrt{37}}{6},\frac{\sqrt{37}+7}{6}\end{bmatrix}$$

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.

$$3x^{2}-7x+1=0$$

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 3 for a, -7 for b, and 1 for c in the quadratic formula.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 1}}{2\times 3}$$

Do the calculations.

$$x=\frac{7±\sqrt{37}}{6}$$

Solve the equation x=\frac{7±\sqrt{37}}{6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{37}+7}{6} x=\frac{7-\sqrt{37}}{6}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\frac{\sqrt{37}+7}{6}\right)\left(x-\frac{7-\sqrt{37}}{6}\right)\leq 0$$

For the product to be ≤0, one of the values x-\frac{\sqrt{37}+7}{6} and x-\frac{7-\sqrt{37}}{6} has to be ≥0 and the other has to be ≤0. Consider the case when x-\frac{\sqrt{37}+7}{6}\geq 0 and x-\frac{7-\sqrt{37}}{6}\leq 0.

$$x-\frac{\sqrt{37}+7}{6}\geq 0 x-\frac{7-\sqrt{37}}{6}\leq 0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{\sqrt{37}+7}{6}\leq 0 and x-\frac{7-\sqrt{37}}{6}\geq 0.

$$x-\frac{7-\sqrt{37}}{6}\geq 0 x-\frac{\sqrt{37}+7}{6}\leq 0$$

The solution satisfying both inequalities is x\in \left[\frac{7-\sqrt{37}}{6},\frac{\sqrt{37}+7}{6}\right].

$$x\in \begin{bmatrix}\frac{7-\sqrt{37}}{6},\frac{\sqrt{37}+7}{6}\end{bmatrix}$$

The final solution is the union of the obtained solutions.

$$x\in \begin{bmatrix}\frac{7-\sqrt{37}}{6},\frac{\sqrt{37}+7}{6}\end{bmatrix}$$