Solve t^2-6t+1geq0

$$t ^ { 2 } - 6 t + 1 \geq 0$$

Solve for t

$$t\in (-\infty,3-2\sqrt{2}]\cup [2\sqrt{2}+3,\infty)$$

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.

$$t^{2}-6t+1=0$$

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 1 for a, -6 for b, and 1 for c in the quadratic formula.

$$t=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$t=\frac{6±4\sqrt{2}}{2}$$

Solve the equation t=\frac{6±4\sqrt{2}}{2} when ± is plus and when ± is minus.

$$t=2\sqrt{2}+3 t=3-2\sqrt{2}$$

Rewrite the inequality by using the obtained solutions.

$$\left(t-\left(2\sqrt{2}+3\right)\right)\left(t-\left(3-2\sqrt{2}\right)\right)\geq 0$$

For the product to be ≥0, t-\left(2\sqrt{2}+3\right) and t-\left(3-2\sqrt{2}\right) have to be both ≤0 or both ≥0. Consider the case when t-\left(2\sqrt{2}+3\right) and t-\left(3-2\sqrt{2}\right) are both ≤0.

$$t-\left(2\sqrt{2}+3\right)\leq 0 t-\left(3-2\sqrt{2}\right)\leq 0$$

The solution satisfying both inequalities is t\leq 3-2\sqrt{2}.

$$t\leq 3-2\sqrt{2}$$

Consider the case when t-\left(2\sqrt{2}+3\right) and t-\left(3-2\sqrt{2}\right) are both ≥0.

$$t-\left(3-2\sqrt{2}\right)\geq 0 t-\left(2\sqrt{2}+3\right)\geq 0$$

The solution satisfying both inequalities is t\geq 2\sqrt{2}+3.

$$t\geq 2\sqrt{2}+3$$

The final solution is the union of the obtained solutions.

$$t\leq 3-2\sqrt{2}\text{; }t\geq 2\sqrt{2}+3$$