I have been thinking of posting some of my Mathematics and Physics lessons on this blog. These are lessons that, I have been using since 1997 during my private tutoring hours, improving over time, mostly for students studying for London General Certificate of Education, Ordinary Level [London GCE O’ Level] and Cambridge International General Certificate of Secondary Education, Ordinary Level [CIE IGCSE O’ Level]. As the very first Mathematics tutorial on this blog, today I am going to show how **quadratic equations can be solved by employing completing the square method**.

### A few assumptions

Before we begin, a few assumptions need to be made regarding the prerequisites for this tutorial as, it so happens that, without prior knowledge of elementary algebra, it will be a bit difficult to grasp ideas that will be conveyed in this lesson. Thus, I will simply assume that, the reader is familiar with and has basic knowledge and understanding of

- Algebra
- The concept of roots, more specifically square roots
- Expansion of binomials and factoring of trinomials using the two identities:
$\left(a+b\right){\phantom{\rule{0ex}{0ex}}}^{2}\equiv a{\phantom{\rule{0ex}{0ex}}}^{2}+2ab+{b}^{2}$

$\left(a-b\right){\phantom{\rule{0ex}{0ex}}}^{2}\equiv a{\phantom{\rule{0ex}{0ex}}}^{2}-2ab+{b}^{2}$

- Quadratic equations including their general form

### Completing the square

Before going in to solving equations, let us first see what completing the square is all about. For that, consider a simple trinomial and see how it can be changed to a perfect square by applying our knowledge of the first identity above. Consider this quadratic expression $${\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{4}\mathbf{x}\mathbf{+}\mathbf{3}$$

Now, how can we turn this into a perfect square, just like the identity? By careful observation it can be seen that, the constant or the last term of a perfect square is always the square of one half the coefficient of the x term. So, it must have 4 as the constant term in this expression. But, it has a 3! How do we change the 3 into 4? Simple! By adding 1. But, by adding 1 to the expression, also changes it to a completely new one changing its meaning. To overcome that, we need to subtract one from it.

So that, the expression retains its original meaning and would now look like this: $${x}^{2}+4x+4-{1}$$Now, ignoring –1, the first three terms of this expression is a perfect square trinomial, which can be expressed as a square of the binomial: $$(x+2){\phantom{\rule{0ex}{0ex}}}^{2}$$

Therefore, the complete expression now looks like this: $$(x+2{)}^{2}-1$$

This process of turning a quadratic expression into a complete square is, what really * completing the square* means.

### Solution of quadratic equations by completing the square

Let us consider the above example by equating it to zero, making it a quadratic equation.

$${\mathit{x}}^{\mathit{2}}\mathbf{+}\mathbf{4}\mathit{x}\mathbf{+}\mathbf{3}\mathbf{=}\mathbf{0}$$

Going through the same steps as above we get

$$(x+2{)}^{2}-1=0$$

Solving for $x$,

$$x=\pm 1-2$$

Giving $x=-1$ and $x=-3$

Note that, one can also get the same results by factoring using the *difference of two squares identity* after completing the square as, 1 is a square number.

### One more example using the second identity

Consider this quadratic equation: $${x}^{2}-6x-7=0$$To complete the square we need to change the constant term to 9. To do that we add 16 to it and subtract 16 to keep the whole expression unchanged giving: $${x}^{2}-6x+9-16=0$$

Taking -16 to RHS and completing square on LHS $$(x-3{)}^{2}=16$$solve for x by taking roots on both sides $$x-3\phantom{\rule{0ex}{0ex}}=\pm \sqrt{16}$$

Therefore, $x=7$ and $x=-1$

### Another example

In the two examples above coefficient of the first term is 1. Let us now see an example with a coefficient greater than 1.

Consider

$$3{x}^{2}-36x-39=0$$This time we will take the constant term to RHS and complete the square on LHS

$$3{x}^{2}-36x=39$$In order to make the coefficient of the first term one, divide the whole equation by 3

$${x}^{2}-12x=13$$Half the coefficient of middle term is -6 squaring it gives 36. This needs to be added to both sides of the equation.

$${x}^{2}-12x+36=13+36$$Simplifying

$${x}^{2}-12x+36=49$$Factoring LHS

$$(x-6{)}^{2}=49$$Solving for x gives, x = 13 and x = -1

That is all about it, for now. Good Luck!

great..!

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