How to derive in a quadratic equation.

Question

I was reading Purplemath's lesson about quadratic equations, and came to the part about deriving the solution to $ x^2 + 6x + 10 = 0$. I understood the part about putting the loose number in the other side and taking half of the coefficient on the x-term (that is, dividing it by two, and keeping the sign), and squaring it, then adding this value to both sides of the equation. But now I don't understand how you do the rest of it. Can someone please explain what you do, how you do it and why?

Answer 1

This is called completing the square. You're using the fact that $x^2 + 2a + a^2 = (x+a)^2$ to solve for $x$, like this:

$$x^2 + 6x + 10 = 0$$ $$x^2 + 6x + 9 = -1$$ $$(x+3)^2 = -1$$ $$x+3 = \pm i$$ $$x = -3 \pm i$$

The key step is that you have a perfect square on both sides of the third equation; you take the square root to get to the fourth equation.

Answer 2

I presume you're asking about the method called "Completing the Square" (good to refer to things by their proper names!).

We can write a general quadratic in the form $x^2 + 2ax - b = 0$.

Here, note that even though your quadratic may not start with a leading coefficient (the one before the $x^2$ term) of one, you can easily divide throughout by it to make it fit that form.

I've also used a coefficient of $2a$ for the $x$ term, and put a negative sign in front of $b$, and these measures are just for simplicity. You'll see why in a second.

Now rearrange that equation:

$x^2 + 2ax = b$

Note that $(x + a)^2 = x^2 + 2ax + a^2$, so the left hand side can be rewritten as: $(x+a)^2 - a^2$.

So $(x+a)^2 - a^2 = b$

Rerrange again:

$(x+a)^2 = b + a^2$

Take square roots of both sides, considering both positive and negative roots:

$x+a = \pm \sqrt{b+a^2}$

Therefore, $x = -a \pm \sqrt{b+a^2}$, which is the solution to your equation.

Let's see how we can apply this to your particular equation. Here $2a = 6$ (or $a = 3$) and $b = -10$.

Hence $x = -3 \pm \sqrt{-10 + 9} = -3 \pm \sqrt{-1} = -3 \pm i$, which are the two complex solutions of your equation.