Solve 6k^2-11k+4>0

$$6 k ^ { 2 } - 11 k + 4 > 0$$

Solve for k

$$k\in \left(-\infty,\frac{1}{2}\right)\cup \left(\frac{4}{3},\infty\right)$$

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.

$$6k^{2}-11k+4=0$$

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 6 for a, -11 for b, and 4 for c in the quadratic formula.

$$k=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 6\times 4}}{2\times 6}$$

Do the calculations.

$$k=\frac{11±5}{12}$$

Solve the equation k=\frac{11±5}{12} when ± is plus and when ± is minus.

$$k=\frac{4}{3} k=\frac{1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$6\left(k-\frac{4}{3}\right)\left(k-\frac{1}{2}\right)>0$$

For the product to be positive, k-\frac{4}{3} and k-\frac{1}{2} have to be both negative or both positive. Consider the case when k-\frac{4}{3} and k-\frac{1}{2} are both negative.

$$k-\frac{4}{3}<0 k-\frac{1}{2}<0$$

The solution satisfying both inequalities is k<\frac{1}{2}.

$$k<\frac{1}{2}$$

Consider the case when k-\frac{4}{3} and k-\frac{1}{2} are both positive.

$$k-\frac{1}{2}>0 k-\frac{4}{3}>0$$

The solution satisfying both inequalities is k>\frac{4}{3}.

$$k>\frac{4}{3}$$

The final solution is the union of the obtained solutions.

$$k<\frac{1}{2}\text{; }k>\frac{4}{3}$$