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In this article, you will learn how to solve any standard **Quadratic Equation** using **the completing the square method**.

We have previously discussed some other methods that can be used to solve a quadratic equation. However, there are generally four methods of solving a standard quadratic equation. These methods are:

- Factorization;
- Completing the square;
- The general quadratic formula(The Almighty Formula);
- Graphical solution.

Each of the four methods above can be used to solve specific types of quadratic equation, but our main focus shall be to apply the “**completing the square method**“.

A standard quadratic equation is of the form ax^2+bx+c=0. So it is considered best practice to make sure that the equation is in its standard form, then apply the following steps below.

#### Using the Completing the Square Method — Steps

A standard quadratic formula is of the form:

ax^2 + bx + c = 0

Where:

a, b are coefficients of x^2 and x respectively;

c is a constant and;

a \ne 0

#### See Also: How to Solve Any Quadratic Equation Using the “Almighty” Formula

Now let’s begin with this **example**:

## Example:

Solve the quadratic equation 2x^2+3x-7=0 using the** completing the square method.**

#### Step 1

Divide the given equation by the coefficient of x^2, (which is a).

This means:

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

For our example, this is:

\frac{2x^2}{2}+\frac{3x}{2}-\frac{7}{2}=\frac{0}{2}

This would simplify into:

x^2+\frac{3x}{2}-\frac{7}{2}=0

(Note that this step is not necessary if a=1)

#### Step 2

Move the constant term, \Large\frac{c}{a} to the other side.

This means:

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

For our example, this is:

x^2+\frac{3x}{2}=\frac{7}{2}

#### Step 3

Add the “**Square**” of “**Half**” of the coeficient of the term with x to both sides.

The term with \large{x} is \Large\frac{bx}{a}. And its coefficient is \Large\frac{b}{a}

The “**Square**” of “**Half**” of this value is \Large(\frac{1}{2}(\frac{b}{a}))^2

Now, this is the same as \Large(\frac{b}{2a})^2

So add this to both sides. This becomes;

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

Further simplified as;

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

Find the L.C.M of the Right hand side and simplify;

x^2+\frac{bx}{a}+(\frac{b}{2a})^2=\frac{-4ac + b^2}{4a^2}

x^2+\frac{bx}{a}+(\frac{b}{2a})^2=\frac{b^2 – 4ac}{4a^2}

For our example, this is:

x^2+\frac{3x}{2}+(\frac{3}{4})^2=\frac{7}{2}+(\frac{3}{4})^2

x^2+\frac{3x}{2}+(\frac{3}{4})^2=\frac{7}{2}+\frac{9}{16}

x^2+\frac{3x}{2}+(\frac{3}{4})^2=\frac{56 + 9}{16}

x^2+\frac{3x}{2}+(\frac{3}{4})^2=\frac{65}{16}

#### Step 4

Complete the square.

To do this, **Factor** the left hand side as the square of a binomial.

The equation then becomes;

(x+\frac{b}{2a})^2=\frac{b^2 – 4ac}{4a^2}

#### Also Recommended: How Smart Are You? Test Yourself With This Simple Mathematics Question

For our example, this is:

(x+\frac{3}{4})^2=\frac{65}{16}

#### Step 5

This is the final step. Take the square root of both sides and solve for \large{x}

x+\frac{b}{2a}=\pm\sqrt\frac{b^2 – 4ac}{4a^2}

x=-\frac{b}{2a}\pm\sqrt\frac{b^2 – 4ac}{4a^2}

x=\frac{-b\pm\sqrt{b^2 – 4ac}}{2a}

For our example, this is:

x+\frac{3}{4}=\pm\sqrt\frac{65}{16}

x=-\frac{3}{4}\pm\sqrt\frac{65}{16}

x=-\frac{3}{4}\pm\frac{\sqrt{65}}{4}

**Final answer in Surd form**:

x=\frac{-3\pm\sqrt{65}}{4}

## ACTIVITIES:

Solve the following quadratic equations using **the completing the square method**

Ans: x=\frac{-7\pm\sqrt{33}}{4}

2. t^2-2t-5=0Ans: t=1\pm\sqrt{6}

**Take Quiz(Coming soon)**