Complete the Square | Methodology, Vertex Form & Problems

Contents hide

In this section, we will learn, “how to complete square of the quadratic equation” and able to find the “solution of the quadratic equation”.

Introduction

Completing the square of any quadratic equation helps us to find the solution of the quadratic equation and gives us a vertex form of any quadratic equation. By vertex form, it is easy to make a graph of any quadratic equation.

First, we see that “what is a quadratic equation?”

The general form of a quadratic equation is,

$ax^{2}+bx+c=0$

In the above expression a (coefficient of $x^{2}$ ) can not be zero. If a = 0, then the equation reduces to a linear equation.

If the solution of the quadratic equation exists, then by completing the square we can convert it into vertex form. By vertex form, we can easily find out the symmetry and turning point (vertex) without plotting the graph.

First, learn how to find out the values from a simple equation. Consider some following examples.

Example: Let the quadratic equation $x^{2}=16$ . Here we want to find the solution of x. We can easily find the solution of x by taking square root on both sides.

$\sqrt{x^{2}}=\pm \sqrt{16}$

Which gives,

x = 4 or x = -4

Always remember whenever we take square roots, we obtained two conclusions, one is in the positive symbol and the other in the negative symbol. The above solution can be written as,

$x=\pm 4$

The procedure to solve $x^{2}=-16$ , is very simple because 16 is a complete square. Since we find the solution 4 and -4, the square of both numbers is 16.

Let us see another example.

Example: Consider the equation $y^{2}=7$ .

To find out the solution of this equation, we take the square root on both sides,

$\sqrt{y^{2}}=\pm \sqrt{7}$

This leads us to the solution,

$y=\sqrt{7}$ or $y=-\sqrt{7}$

In this problem, the right-hand side of the expression is not a perfect square, but still, we can find the solution. Leave the result in actual form. By using a calculator, the results may disturb. Because the calculator gives answers in decimal form.

Now see another example.

Example: Find the solution of $(x-5)^{2}=4$

Here we also start by taking square root on both sides of the equation.

$x-5=\pm 2$

Which mean,

x – 5 = 2 or x – 5 = -2

Here adding 5 on both sides of each expression, leads us to the solution,

x – 5 + 5 = 2 + 5 or x – 5 + 5 = -2 + 5

x = 7 or x = 3

So, the solution is x = 7 and x = 3.

Methodology

We will face two types of the equation,

Coefficient of $x^{2}$ is 1
Coefficient of $x^{2}$ is not 1

First, we discuss the first type, that is coefficient of $x^{2}$ is 1.

Consider, we have to find the solution for $x^{2} + 2 x = 8$ .

In all previous examples, both sides of the equation were completely square.

But, in this equation, the left-hand side is not a complete square. To find out the solution left side must be complete square or in the form of $(x+a)^{2}$ or $(x-a)^{2}$ . To convert the left-hand side into a complete square we go through the following procedure.

First dividing the coefficient of x, by 2
Then the square of the resulting number added on both side
The left-hand side will be summed up into square form.

In the equation $x^{2} + 2 x = 8$ , coefficient of x is 2.

Dividing the coefficient of x by 2,

$\frac{Coefficient\,of\, x}{2}=\frac{2}{2}=1$

Now taking the square of the resulting number,

$(1)^{2}=1$

Adding the resulting number on both sides of the equation,

$x^{2} + 2 x +1= 8+1$

As we know $(x+a)^{2}=x^{2}+2ax+a^{2}$ , comparing the left-hand side, we get,

$(x+1)^{2}=9$

Now the left-hand side is complete square, which can easily be solved by taking square root on both sides of the equation.

Taking square root on both sides of the equation,

$x+1=\pm 9$

$x+1=\pm 3$

The solution of $x^{2} + 2 x= 8$ is,

x+1=3 or x+1=-3

x=3-1 or x= -3-1

x=2 or x= -4

Now we will see how to solve quadratic equations by completing the square when the coefficient of $x^{2}$ is not 1.

We follow the following steps,

If the coefficient of $x^{2}$ is not one, then make the coefficient one of $x^{2}$ by taking common from the terms including $x^{2}$ and x, not from the constant term or dividing the whole equation by coefficient of $x^{2}$ .
After it, the whole procedure is the same as we do when the coefficient of $x^{2}$ is one.

Let us take an example to see how the method work.

Example: Solve the quadratic equation $2x^{2}-8x+6=0$ by completing the square.

In this equation coefficient of $x^{2}$ is not one. So first make the coefficient one by dividing the coefficient of $x^{2}$ to the whole equation.

Dividing the equation by 2, since the coefficient of $x^{2}$ is 2.

$\frac{2x^{2}}{2}-\frac{8x}{2}+\frac{6}{2}=0$

We get,

$x^{2}-4x+3=0$

Now according to our last procedure, let the coefficient of x, which is 4.

Dividing 4 by 2 we get,

$\frac{-4}{2}=-2$

Now take the square of -2 and add on both side of the resulting equation,

$x^{2}-4x+(-2)^{2}+3=(-2)^{2}$

Rearranging the equation,

$x^{2}-2(2)x+2^{2}=-3+4$

Recall the formula $(x-a)^{2}=x^{2}-2ax+a^{2}$ , we get,

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

Now, this is a simple equation, which can be solved easily by taking square root on both sides of the equation.

By taking square root we will get,

$x-2=\pm 1$

Evaluate the value of x, as

x – 2 = 1 or x – 2 = -1

x = 1 + 2 or x = -1 + 2

x = 3 or x = 1

Which is the required solution.

Vertex Form

Through the procedure of completing the square, we can easily develop the vertex form of a quadratic equation.

What is vertex form?

Vertex form of the quadratic equation is a special shape, through which we can easily see that what is the symmetry of the graph of the quadratic equation and also be able to see the turning point on the graph.

The standard form of the vertex form of any quadratic equation is,

$y=a(x-h)^{2}+k$

Where (h,k) is said to be vertex or turning point of the graph and a shows the direction of the graph, if a is positive then the graph will be upward and if a is negative then the graph will be downward.

As we know the shape of the graph of any quadratic equation is always parabolic shape.

In this form, h is the point of x-axis where the graph has symmetry and k is the minimum or maximum value on the graph.

Example: Convert the quadratic equation $y=3x^{2}+12x+3$ in vertex form.

Since the given equation is not in a perfect square, to convert the equation incomplete square, we follow the method the same method we described above.

Let

$y=3x^{2}+12x+3$

Taking common 3 on the right-hand side,

$y=3(x^{2}+4x+1)$

Divide the whole equation by 3. We get,

$\frac{y}{3}=x^{2}+4x+1$

Take the coefficient of x from the right-hand-side, which is 4. Now divide it by 2,

$\frac{4}{2}=2$

Now add the square of the resulting number on both side of the equation, we get

$\frac{y}{3}+2^{2}=x^{2}+4x+2^{2}+1$

In the above equation the expression $x^{2}+4x+4$ is the complete square of (x+2), writing this expression in term of the square,

$\frac{y}{3}+4=(x+2)^{2}+1$

Subtracting 4 on both sides,

$\frac{y}{3}+4-4=(x+2)^{2}+1-4$

$\frac{y}{3}=(x+2)^{2}-3$

Multiplying by 3,

$y=3(x+2)^{2}-9$

Which is the required vertex form of the equation.

Comparing $y=3(x+2)^{2}-9$

With

$y=a(x-h)^{2}+k$

We get, h=-2, k=-9 and a=3. So the vertex is (-2,-9) which means the graph has the symmetry at x=-2, the minimum value of the graph is -9, and since a is positive so the graph has an upward direction. As shown below.

In the graph, we can see that the turning point of the graph is (-2,-9) and the direction of the graph is upward, and the minimum value of the graph is -9.

Problems:

Problem 1: Solve the quadratic equation $5x^{2}+4x-2=0$ by completing the square.

We have to find the solution of the given quadratic equation by completing the square. To find the solution of the given equation, we first convert the right-hand side into a square form. Given equation is,

$5x^{2}+4x-2=0$

Coefficient of $x^{2}$ is not one. First, we make the coefficient of $x^{2}$ equal to one. To make the coefficient equals to one, divide the whole equation by the coefficient of $x^{2}$ . Coefficient of $x^{2}$ is “5”. So dividing the whole equation by “5” we get,

$\frac{5x^{2}}{5}+\frac{4x}{5}-\frac{2}{5}=0$

Which reduce to,

$x^{2}+\frac{4x}{5}-\frac{2}{5}=0$

Our next step is, we take the coefficient of x, and divide it by “2”, then summed up the square of the resulting number on both sides of the equation. Since the coefficient of x is 4/5.

Dividing the coefficient of x by “2”, we get,

$\frac{\frac{4}{5}}{2}=\frac{4}{5}(\frac{1}{2})=\frac{2}{5}$

The resulting number is 2/5, now take the square of the resulting number and add both sides of the equation,

$x^{2}+\frac{4x}{5}+(\frac{2}{5})^{2}-\frac{2}{5}=0+(\frac{2}{5})^{2}$

Rearranging the equation,

$x^{2}+2(\frac{2}{5})x+(\frac{2}{5})^{2}=\frac{4}{25}+\frac{2}{5}$

Now left-hand side is complete square,

$(x+\frac{2}{5})^{2}=\frac{4}{25}+\frac{2}{5}$

Simplifying the right-hand side of the equation,

$(x+\frac{2}{5})^{2}=\frac{4+10}{25}=\frac{14}{25}$

Taking square root on both sides we get,

$x+\frac{2}{5}=\pm \frac{\sqrt{14}}{5}$

Shifting 2/5 from the left side to the right side, we get

$x=-\frac{2}{5}\pm \frac{\sqrt{14}}{5}=\frac{14}{25}$

$x=\frac{-2\pm \sqrt{14}}{5}$

This is the required solution of the quadratic equation. Which can also be evaluated by a calculator to find out the values in decimal form.

Problem 2: Write the expression $4x^{2}-4x-4$ in the complete square.

To make the complete square of the given expression, we make the coefficient one of $x^{2}$ . To make the coefficient one, take common 4 from the whole expression.

$4(x^{2}-x-1)$

Now, go to the next step. Which is dividing the coefficient of x, by 2. The coefficient of x is 1.

Dividing 1 by 2 we get, 1/2.

Adding and subtracting the square of the resulting number. We get,

$4(x^{2}-x+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}-1)$

Rearranging the expression,

$4(x^{2}-2(\frac{1}{2})x+(\frac{1}{2})^{2}-\frac{1}{4}-1)$

Writing in term of the square,

$4((x-\frac{1}{2})^{2}-\frac{1}{4}-1)$

Adding the constants,

$4((x-\frac{1}{2})^{2}+\frac{-1-4}{-4})$

Here is the final form in term of the square,

$4((x-\frac{1}{2})^{2}-\frac{5}{4})$

Or it can also be written as, by multiplying 4 inside,

$4(x-\frac{1}{2})^{2}-5$

Problem 3: convert the quadratic equation $y=x^{2}-x+9$ in vertex form, write the axis of symmetry and coordinates of the vertex.

Here we will first complete the square of the equation. To convert into vertex form we should have to complete the square.

In this quadratic equation coefficient of $x^{2}$ is one, so we will take start from the second step.

In the procedure of completing the square, our second step is dividing the coefficient of x by two. In our present equation, the coefficient of x, is one. So, dividing 1 by 2, we will get 1/2.

Adding and subtracting square of (1/2) we get,

$y=x^{2}-x+(\frac{1}{2})^{2}-(\frac{1}{2})^{2}+9$

Rearranging the expression,

$y=x^{2}-2(\frac{1}{2})x+(\frac{1}{2})^{2}-\frac{1}{4}+9$

Writing in the term of the complete square and adding the constants. We get

$y=(x-\frac{1}{2})^{2}+\frac{-1+36}{4}$

$y=(x-\frac{1}{2})^{2}+\frac{35}{4}$

Here is the vertex form of the given equation. Comparing it by standard vertex form,

$y=a(x-h)^{2}+k$

We get, $h=\frac{1}{2}$ , $k=\frac{35}{4}$ and a = 1. So, the coordinates of the vertex are $(\frac{1}{2},\frac{35}{4})$ , and symmetry of axis is x = 1/2.

Summary:

In this article of complete the square, we learn how to turn the expression into the complete square. This is very useful in the solution of analytical geometry. We use this method to find out the main points of the parabolic graphs.

Source:

AS Level Mathematics Textbook by Aaran Karia
Pure mathematics 1 by Sophie Goldie