Solve 36k^2-4k(k+8)<0

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

Solve for k

$$k\in \left(0,1\right)$$

Solution Steps

Multiply -1 and 4 to get -4.

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

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

$$36k^{2}-4k^{2}-32k<0$$

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

$$32k^{2}-32k<0$$

Factor out k.

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

For the product to be negative, k and k-1 have to be of the opposite signs. Consider the case when k is positive and k-1 is negative.

$$k>0 k-1<0$$

The solution satisfying both inequalities is k\in \left(0,1\right).

$$k\in \left(0,1\right)$$

Consider the case when k-1 is positive and k is negative.

$$k-1>0 k<0$$

This is false for any k.

$$k\in \emptyset$$

The final solution is the union of the obtained solutions.

$$k\in \left(0,1\right)$$