Solve (k-3)^2-4(1)(9)<0

$$( k - 3 ) ^ { 2 } - 4 ( 1 ) ( 9 ) < 0$$

Solve for k

$$k\in \left(-3,9\right)$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(k-3\right)^{2}.

$$k^{2}-6k+9-4\times 1\times 9<0$$

Multiply 4 and 1 to get 4.

$$k^{2}-6k+9-4\times 9<0$$

Multiply 4 and 9 to get 36.

$$k^{2}-6k+9-36<0$$

Subtract 36 from 9 to get -27.

$$k^{2}-6k-27<0$$

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.

$$k^{2}-6k-27=0$$

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 -27 for c in the quadratic formula.

$$k=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 1\left(-27\right)}}{2}$$

Do the calculations.

$$k=\frac{6±12}{2}$$

Solve the equation k=\frac{6±12}{2} when ± is plus and when ± is minus.

$$k=9 k=-3$$

Rewrite the inequality by using the obtained solutions.

$$\left(k-9\right)\left(k+3\right)<0$$

For the product to be negative, k-9 and k+3 have to be of the opposite signs. Consider the case when k-9 is positive and k+3 is negative.

$$k-9>0 k+3<0$$

This is false for any k.

$$k\in \emptyset$$

Consider the case when k+3 is positive and k-9 is negative.

$$k+3>0 k-9<0$$

The solution satisfying both inequalities is k\in \left(-3,9\right).

$$k\in \left(-3,9\right)$$

The final solution is the union of the obtained solutions.

$$k\in \left(-3,9\right)$$