# Explain how solving quadratics by completing the square works,

I am trying to understand completing the square.

The most helpful site I have found in Maths is Fun.

I don't know what the policy is on links, but I believe it might help tremendously to use the page I'm reading because it does a lot of the work.

Explanation of completing the square

It begins with an explanation using $x^2 + bx$.

What I understand from the explanation is $x^2 + bx = (x + \frac{b}{2})^2 - (\frac{b}{2})^2$. Correct?

It also shows how $x^2 + bx$ can almost be rearranged into a square, and that square can be completed with the addition of $(\frac{b}{2})^2$. Right so far?

It then goes on to show completing the square algebraically using both the summing of three components or the squaring of two components summed. So far so good?

It adds: To make everything balance, if you are to add that last corner, then you must take it away again. At this point I don't know if I'm following. I'm thinking "Ok, but how does this help us at all?".

It then goes into an example:

$x^2 + 6x + 7$

The next two steps are the ones I don't fully understand.

I can see we're adding half b squared, as in the example, but this isn't the example anymore, it's a quadratic equation now.

It shows how to add in that corner piece, then take it away again at the end. What was b in the generalised form is now 6 in the example case.

The simplification, I have never understood in any example, so if I'm right so far, this is where I need help.

The $7 - 9$ part, fine. It just did the calculation with real numbers.

For the $x^2 + 6x + (\frac{6}{2})^2$ half though, I don't see how the simplification happens.

Oh yes, I do. I've just seen it.

Just as well I complete this post for anybody to comment on or correct, because it's a useful subject for anybody moving towards calculus.

The $x^2 + bx$ is where we started, with those helpful illustrative diagrams showing why it all works. We saw that when you add $\frac{b}{2}^2$ you get a perfect square. We also saw that square can also be expressed $(x + \frac{b}{2})^2$

OK, I have now followed the example through to the finish.

I'm still a bit lost on how this solves quadratics because we now have the quadratic unexpanded, into its binomial form, but so what? What's x? I now have to solve the equation given for x?

• Well, you need to have an equation. $x^2 + bx = \text{something}$. Right? – fleablood Nov 4 '15 at 4:05
• Note that you don't need an equation to complete the square. This web page shows you how to factor by completing the square: mathuprising.comlu.com/special-products-case-study.html – John Joy Nov 4 '15 at 13:40
• Examples 1 and 2 at the bottom of the Math Is Fun page show you how to solve a quadratic equation by completing the square. Do you understand them? If not, what specific difficulties are you having? – N. F. Taussig Nov 10 '15 at 10:57

## 3 Answers

You need an equation. $x^2 + bx =\text{something}$.

$x^2 + bx = c$

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

Now $(x + \frac b 2)^2 = x^2 + bx + \frac {b^2}{4}$ So

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

$x + \frac b 2 = \pm \sqrt{c +\frac {b^2}{4} }$ So

$x = \frac b 2 \pm \sqrt{c +\frac {b^2}{4} }$ We're done but to simplify

$x = \frac b 2 \pm \sqrt{\frac {4c}4 +\frac {b^2}{4} }=\frac {b \pm \sqrt{4c + b^2}}{2}$

You should be thinking this looks a lot like the quadradic equation.

For perfect square you have to add $($1/2$.coefficient of x)$^2 on both the sides of the equation . The equation should be of the form $x^2$+bx. This is how to convert a equation into a perfect square.(more precise the LHS).

I find the explanation provided in your link extremely convoluted and leads the reader to believe the “adjusting constant” needs to be added where it need to be subtracted. I can see that in your statement above “ ... square can be completed with the addition of ...” when it should be “ ... subtraction of ...”. The method I was taught to use is separate out the constant term and deal with the terms in $x$ and $x^2$. Factorise $x^2 + bx + c \equiv (x + \frac{b}{2})^2 –(\frac{b}{2})^2 + c$. Then tidy up the constants. Also: $ax^2 + bx + c \equiv a(x^2 + \frac{b}{a}x) +c \equiv a\Big\{(x + \frac{b}{2a})^2 –(\frac{b}{2a})^2\Big\} + c \equiv a(x + \frac{b}{2a})^2 –a(\frac{b}{2a})^2 + c$ and again tidy up the constants.

Also

In a solution above there is an error in the sign of $\frac{b}{2}$ as it is moved from LHS to RHS. I corrected that but the correction has been removed.