dear,
anybody knws how is the easiest way to completing square? or a more simple way?
ex : x^2 - 4x - 5 = 0
thank you
Note that if you expand $\displaystyle (x + a)^2$ you get $\displaystyle x^2 + 2ax + a^2$. Now look at the 'a' value in the 2ax term. That is your 'a' value in the $\displaystyle (x + a)^2$ expression.
So if you have a equation such as $\displaystyle x^2 + 6x + 6$... This can be re-written as $\displaystyle x^2 + 2 \cdot 3x + 6$ and hence 3 is your 'a' value. So in this example completing the square would be...
$\displaystyle x^2 + 6x + 6 = (x + 3)^2 - 3$ since expanding $\displaystyle (x + 3)^2$ gives $\displaystyle x^2 + 6x + 9$ so you need to take away a 3 from it to match your original equation.
Basically to generalize... If you have an equation of the form $\displaystyle x^2 + nx + m$ where n and m are some numbers, you should divide n by 2 to get the 'a' value in the equation $\displaystyle (x + a)^2$.
So lets do your example...
$\displaystyle x^2 - 4x - 5 = 0$
Your n value is -4. Divide this by 2 to get -2. This is your 'a' value. Put this into the equation $\displaystyle (x + a)^2$ to get... $\displaystyle (x - 2)^2$.
Now expand this... $\displaystyle (x - 2)^2 = x^2 - 4x + 4$. Since this is different from you original equation you need to subtract 9 to turn the +4 into a -9...
So your answer is! $\displaystyle (x - 2)^2 - 9$.
Hello janeyandtazCompleting the square is a bit tricky to explain, but you may find you can follow the technique I've described here, where I've broken it down to 6 steps in my answer to Question 1. (You'll see that I've given an example where the coefficient of $\displaystyle x$ is not equal to $\displaystyle 1$.)
Grandad