Solve 36k^2-4k(k+8)leq0

$$36 k ^ { 2 } - 4 k ( k + 8 ) \leq 0$$

Solve for k

$$k\in \begin{bmatrix}0,1\end{bmatrix}$$

Solution Steps

Multiply -1 and 4 to get -4.

$$36k^{2}-4k\left(k+8\right)\leq 0$$

Use the distributive property to multiply -4k by k+8.

$$36k^{2}-4k^{2}-32k\leq 0$$

Combine 36k^{2} and -4k^{2} to get 32k^{2}.

$$32k^{2}-32k\leq 0$$

Factor out k.

$$32k\left(k-1\right)\leq 0$$

For the product to be ≤0, one of the values k and k-1 has to be ≥0 and the other has to be ≤0. Consider the case when k\geq 0 and k-1\leq 0.

$$k\geq 0 k-1\leq 0$$

The solution satisfying both inequalities is k\in \left[0,1\right].

$$k\in \begin{bmatrix}0,1\end{bmatrix}$$

Consider the case when k\leq 0 and k-1\geq 0.

$$k-1\geq 0 k\leq 0$$

This is false for any k.

$$k\in \emptyset$$

The final solution is the union of the obtained solutions.

$$k\in \begin{bmatrix}0,1\end{bmatrix}$$