Questions: Solve the equation by the quadratic formula. Give only real number solutions. (x-3)(x+9)=-16

Solution Steps

To solve the equation \((x-3)(x+9)=-16\), first expand the left side to form a standard quadratic equation. Then, rearrange the equation to set it to zero. Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find the real solutions, if they exist.

Starting with the equation \((x-3)(x+9) = -16\), we first expand the left side: \[ x^2 + 9x - 3x - 27 = -16 \] This simplifies to: \[ x^2 + 6x - 27 = -16 \] Next, we rearrange the equation to set it to zero: \[ x^2 + 6x - 11 = 0 \]

For the quadratic equation \(ax^2 + bx + c = 0\), we identify \(a = 1\), \(b = 6\), and \(c = -11\). The discriminant \(D\) is calculated as: \[ D = b^2 - 4ac = 6^2 - 4 \cdot 1 \cdot (-11) = 36 + 44 = 80 \]

Since the discriminant \(D = 80\) is positive, this indicates that there are two distinct real solutions.

Using the quadratic formula \(x = \frac{-b \pm \sqrt{D}}{2a}\), we find the solutions: \[ x = \frac{-6 \pm \sqrt{80}}{2 \cdot 1} = \frac{-6 \pm 4\sqrt{5}}{2} = -3 \pm 2\sqrt{5} \] Thus, the two solutions are: \[ x_1 = -3 + 2\sqrt{5}, \quad x_2 = -3 - 2\sqrt{5} \]

Final Answer