Solve for $x$:
$x^2 + 2x = -20$
The problem asks to solve the quadratic equation \(x^2 + 2x = -20\) for \(x\). First, we rearrange the equation into the standard form \(ax^2 + bx + c = 0\) by adding 20 to both sides:
\[x^2 + 2x + 20 = 0\]
Here, the coefficients are \(a = 1\), \(b = 2\), and \(c = 20\). Since this is a quadratic equation, we can use the quadratic formula to find the roots. The quadratic formula is given by:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Substitute the values \(a = 1\), \(b = 2\), and \(c = 20\) into the quadratic formula:
\[x = \frac{-2 \pm \sqrt{2^2 - 4(1)(20)}}{2(1)}\]
Calculate the discriminant:
\[b^2 - 4ac = 2^2 - 4(1)(20) = 4 - 80 = -76\]
Since the discriminant is negative, the solutions will involve complex numbers. We can express the square root of \(-76\) using the imaginary unit \(i\), where \(i = \sqrt{-1}\):
\[\sqrt{-76} = \sqrt{76} \times \sqrt{-1} = \sqrt{4 \times 19} \times i = 2\sqrt{19}i\]
Now, substitute this back into the expression for \(x\):
\[x = \frac{-2 \pm 2\sqrt{19}i}{2}\]
Divide both terms in the numerator by the denominator:
\[x = -1 \pm \sqrt{19}i\]
Alternatively, we can complete the square to solve for \(x\). Start with \(x^2 + 2x = -20\). To make the left side a perfect square trinomial, we add \(\left(\frac{b}{2}\right)^2 = \left(\frac{2}{2}\right)^2 = 1^2 = 1\) to both sides:
\[x^2 + 2x + 1 = -20 + 1\]
\[(x + 1)^2 = -19\]
Take the square root of both sides:
\[x + 1 = \pm\sqrt{-19}\]
Since \(\sqrt{-19} = \sqrt{19}i\), we have:
\[x + 1 = \pm\sqrt{19}i\]
Subtract 1 from both sides to isolate \(x\):
\[x = -1 \pm \sqrt{19}i\]
Both methods yield the same result, confirming the values of \(x\).
The solutions to the quadratic equation \(x^2 + 2x = -20\) are \(x = -1 + \sqrt{19}i\) and \(x = -1 - \sqrt{19}i\). The final answer is \(\boxed{-1 \pm \sqrt{19}i}\).
Quadratic equations with complex roots appear frequently in engineering and physics, particularly in the study of damped oscillations. For instance, if you are modeling the movement of a mechanical system or the behavior of an electrical circuit where energy is dissipating (damping), the characteristic equation often results in complex roots. These roots tell us not only how fast the system returns to equilibrium but also the frequency at which it oscillates as it approaches that equilibrium point. Understanding how to solve these equations is essential for predicting the stability and performance of such systems.