As we know, solving the quadratic equation by using the Quadratic formula is quite easy. But when it is said to solve the Quadratic equation by completing the square method then students get a little bit confused. If you are included in those who are finding Quadratic equation solvers or to find the quadratic function calculator to solve quadratic equations by completing square method, don’t worry we will completely guide you how you will solve these quadratic equations.

### Completing The Square Method To Solve The Quadratic Equation

In completing the square method, we follow the following steps:

#### Step# 1 Transposition The Equation:

First transpose the equation if needed.
OR write the quadratic equation in standard form.

#### Step # 2 Make The Coefficient Of X Is Equal To 1:

Divide the exponent of the x by the whole equation to make the coefficient of x is equal to 1.

#### Step # 3 Make A Perfect Square:

Add the value $$(½ × coefficient of x)^2$$ to each side of the equation to make it a perfect square.

#### Step # 4: Simplify The Equation:

Convert the left side of the equation into square form and simplify the values of the right side of the equation.

#### Step # 5 Take The Square Root:

Take the square root of the both sides of the equation and solve the equation with basic rules like add and subtract then finally after solving the equation we will get the value of x or required solution sets.

Let’s understand the completing the square method to solve the quadratic equations by the following examples:

#### Example# 1:

Solve the equation by completing the square method.

$$X^2 – 2x = 2$$

Solution:

Here we skip the first and second step because the equation is already in standard form and there is no need to make the exponents of x equal to 1. Now let’s start other steps:

Add $$(\frac{1}{2} × b)^2 = (\frac{1}{2} × -2)^2 = (-1)^2 = 1$$ on the both sides of the equation,

$$X^2 – 2x + 1 = 2 + 1$$

$$(x-1)^2 = 3$$

Now taking square root on both sides of the equation,

$$\sqrt{(x-1)^2} = \sqrt{3}$$

$$x-1 = \pm \sqrt{3}$$

$$x = 1 \pm \sqrt{3}$$

Solution sets {$$1 – \sqrt{3}$$, $$1 + \sqrt{3}$$} Answer.

#### Example# 2:

Solve the equation by completing the square method:

$$12 = 4x + 5x^2$$

#### Solution:

By transposition the equation we have,

$$5x^2 + 4x = 12$$

Now divide the exponent of x by the whole equation to make it equal to 1,

$$\dfrac{5x^2}{5} + \dfrac{4x}{5} = \dfrac{12}{5}$$

$$x^2 + \dfrac{4x}{5} = \dfrac{12}{5}$$

Add $$(\dfrac{1}{2} × coefficient of x)^2$$

We have $$(\dfrac{1}{2} × \dfrac{4}{5})^2 = (\dfrac{4}{10})^2 = 16/100$$

as $$\dfrac{16}{100}$$ cuts by the table of $$4 = \dfrac{4}{25}$$

Add $$\dfrac{4}{25}$$ on both sides of the equation,

$$x^2 + \dfrac{4}{25}x + \dfrac{4}{25} = \dfrac{12}{5} + \dfrac{4}{25}$$

Note:

At the left side of the equation the square term is $$x^2 + \dfrac{4}{25}$$ we can write it as

$$(x + \dfrac{2}{5})^2$$ Now it is as $$(a + b)^2$$ so, we apply the formula $$a^2 + 2ab + b^2$$ and here $$a = x$$ & $$b = \dfrac{2}{5}$$.

$$x^2 + 2(x)(\dfrac{2}{5}) + (\dfrac{2}{5})^2 = \dfrac{12}{5} + \dfrac{4}{25}$$

$$(x + \dfrac{2}{5})^2 = \dfrac{12}{5} + \dfrac{4}{25}$$

$$(x + \dfrac{2}{5})^2 = \dfrac{60 + 4}{25}$$

$$(x + \dfrac{2}{5})^2 = \dfrac{64}{25}$$

Now taking square root on the both sides of the equation,

$$\sqrt{(x + \dfrac{2}{5})^2} = \sqrt{\dfrac{64}{25}}$$

$$x + \dfrac{2}{5} = \pm \dfrac{8}{5}$$

$$x + \dfrac{2}{5} = \dfrac{8}{5}$$ OR $$x + \dfrac{2}{5} = -\dfrac{8}{5}$$

$$x = – \dfrac{2}{5} + \dfrac{8}{5}$$ OR $$x = – \dfrac{2}{5} -\dfrac{8}{5}$$

$$x = \dfrac{-2 + 8}{5}$$ OR $$x = \dfrac{-2 – 8}{5}$$

$$x = \dfrac{6}{5}$$ OR $$x = \dfrac{-10}{5}$$

$$x = 1.2$$ OR $$x = -2$$

So, the solution sets {1.2, -2} Answer.