Question

I am having trouble understanding the concept of completing the square in math. Could someone please assist me on how to do so?

Answer 1

"Completing the square" is the process used to derive the quadratic  formula for the general quadratic ax^2+bx+c=0.  Suppose you did not know the value of a,b, or c of the quadratic...

ax^2+bx+c=0  You need a leading coefficient of one for the process to work, so you divide the whole equation by a

x^2+bx/a+c/a=0  now you move the constant to the other side of the equation

x^2+bx/a=-c/a  now you halve the linear coefficient, square that, then add that value to both sides, ie, (b/(2a))^2=b^2/(4a^2)...

x^2+bx/a+b^2/(4a^2)=b^2/(4a^2)-c/a  now the left side is a perfect square...

(x+b/(2a))^2=(b^2-4ac)/(4a^2)  now take the square root of both sides

x+b/(2a)=±√(b^2-4ac)/(2a)  now subtract b/(2a) from both sides

x=(-b±√(b^2-4ac))/(2a)

It is actually much simpler keeping track of everything when using known values for a,b, and c, but the above explains the actual process used to create the quadratic formula, which the above solution is. :)

Answer 2

If you have a quadratic equation in the form $ax^2 + bx + c = 0$ you can complete the square with the following instructions.

1. If the coefficient of $ax^2$ is not equal to 1, then you will need to divide the whole equation by a. Or simply divide the whole equation by a.

$ax^2 + bx + c = 0 \\ \\ \frac{ax^2}{a} + \frac{bx}{a} + \frac{c}{a} = 0 \\ \\ x^2 + \frac{bx}{a} + \frac{c}{a} = 0$

2. Then get all the x terms on one side of the equation. We do this by subtracting $\frac{c}{a}$ from both sides of the equation.

$x^2 + \frac{bx}{a} + \frac{c}{a} - \frac{c}{a} = 0 - \frac{c}{a} \\ \\ x^2 + \frac{bx}{a} = - \frac{c}{a}$

3. Finally, take the half of the coefficient of $\frac{bx}{a}$, square it and add it to both sides of the equation.

$\frac{bx}{a} = \frac{b}{a}x \\ \\ \frac{b}{2a} \\ \\ (\frac{b}{2a})^2 = \frac{b^2}{4a^2} \\ \\ x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} = \frac{c}{a} + \frac{b^2}{4a^2}$

I hope that helped!