Question

Can someone please help me with Solving Quadratic Equations with Complex Solutions????

Answer 1

Post your equation

Answer 2

Solve x2 + 2x + 9 = 0.

Answer 3

@SithsAndGiggles

Answer 4

You can use the quadratic formula directly:
\[x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\]
with \(a=1,b=2,c=9\), then simplify the results.

You can also complete the square:
\[\begin{align*}
x^2+2x+9&=x^2+2x+1+8\\
0&=(x+1)^2+8
\end{align*}\]
This makes it somewhat easier to isolate \(x\) if you're not comfortable with the previous formula (though you should definitely be comfortable with it, it's pretty important).

A slightly harder way would be to factorize. This will give you the roots immediately. With a general equation, however, you might want to try a system of equations approach. We have
\[ax^2+bx+c=x^2+2x+9\]
and we want something of the form \(a(x-x_1)(x-x_2)\), where \(x_1\) and \(x_2\) are the roots/solutions. Expanding this, we get
\[ax^2-ax_1x-ax_2x+ax_1x_2\]
Matching up terms, we have
\[\begin{align*}\color{red}ax^2+\color{blue}{(-ax_1-ax_2)}x+\color{green}{ax_1x_2}&=\color{red}{1}x^2+\color{blue}2x+\color{green}9\end{align*}\]
so you have
\[\begin{cases}
a=1\\
-ax_1-ax_2=2\\
ax_1x_2=9
\end{cases}~~\iff~~
\begin{cases}
-x_1-x_2=2\\
x_1x_2=9
\end{cases}\]
and from here you solve for \(x_1\) and \(x_2\).