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Last modified on August 3rd, 2023

Completing the square is a way of rearranging quadratic equations from the general form ax^{2} + bx + c = 0 to the vertex form a(x – h)^{2} + k = 0. It is written as a(x + m)^{2} + n, such that the left side is a perfect square trinomial.

Comparing this equation with the general form, we get:

- ${m=\dfrac{b}{2a}}$
- ${n=c-\dfrac{b^{2}}{4a}}$

It is a form used to solve quadratic equations that is also useful in graphing and analyzing a quadratic function and finding its minimum or maximum values.

For example, the quadratic equation x^{2} – 4x + 4 = 0 can also be written as (x – 2)^{2}. Although the two equations are equivalent, the form (x – 2)^{2} is sometimes better to work with in some situations.

Let us try to understand the concept using the concept of geometry.

In a quadratic equation ax^{2} + bx + c, we will arrange the expression in the form of a perfect square trinomial. Here, x comes twice, which makes it tough to solve. We will simplify the equation further using the concept of geometry as shown below:

We will start by dividing all terms by a, the coefficient of x^{2}

Now, let us divide the equation into a square of side x and a rectangle of length ${\dfrac{b}{a}}$ and breadth x as shown:

Thus by adding ${\left( \dfrac{b}{2a}\right) ^{2}}$to the equation ${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$we can complete the square.

**Step 1: **Dividing all terms by a, the coefficient of x^{2}

${x^{2}+\dfrac{b}{a}x+\dfrac{c}{a}=0}$

**Step 2**: Rewriting the equation such that the constant term ‘c’ is the only term on the right side of the equation

=>${x^{2}+\dfrac{b}{a}x=-\dfrac{c}{a}}$

**Step 3**:Adding the square of half the coefficient of the x -term, ${\left( \dfrac{b}{2a}\right) ^{2}}$ to both sides of the equation.

**Step 4**:Factoring the left side as the square of the binomial

=>${x^{2}+\dfrac{b}{a}x+\left( \dfrac{b}{2a}\right) ^{2}=-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}}$

= ${\left( x+\dfrac{b}{2a}\right) ^{2}= -\dfrac{c}{a}+\dfrac{b}{( 2a}) ^{2}}$

This equation is in the vertex form a(x – h)^{2} + k. From here, we can find the vertex

**Step 5**: Taking the square root of both sides of the equation

=>${\sqrt{\left( x+\dfrac{b}{2a}\right) ^{2}}=\sqrt{-\dfrac{c}{a}+\begin{bmatrix} – \\ 20 \end{bmatrix}^{2}}}$

**Step 6**: Subtracting the number that remains on the left side of the equation to find x

=>${x+\dfrac{b}{2a}-\dfrac{b}{2a}=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$

=>${x=\sqrt{-\dfrac{c}{a}+\left( \dfrac{b}{2a}\right) ^{2}-}\dfrac{b}{2a}}$

By completing the square, let us solve the quadratic equation x^{2} + 4x + 1 = 0.

**Step 1**: Here, this step is not needed as a = 1

**Step 2**:Moving the constant term to the right side of the equation

=>x^{2} + 4x = -1

**Step 3**:: Adding the square of half the coefficient of the x -term to both sides of the equation.

${\left( \dfrac{4}{2}\right) ^{2}=\left( 2\right) ^{2}=4}$

=>x^{2} + 4x + 4 = -1 + 4

**Step 4**:Factoring the left side as the square of the binomial

=>(x + 2)^{2} = 3

This gives us the vertex of the parabola corresponding to the given equation as (-2, -3)

**Step 5**: Taking the square root on both sides of the equation

=>x + 2 = ±√3 = ±1.73

**Step 6**: Subtracting 2 from both sides of the equation

=> x = (-2 + √3, -2 – √3)

**Find the value of c in the given quadratic equation x ^{2} + 9x + c that completes the square.**

Solution:

Given the quadratic equation, x^{2} + 9x + c, here b = 9

As we know,

x^{2} + 9x + c

=> ${c=\left( \dfrac{b}{2}\right) ^{2}}$

=> ${c=\left( \dfrac{9}{2}\right) ^{2}}$

=> ${c=\dfrac{81}{4}}$

So, the value of c that makes the quadratic equation a perfect square trinomial is ${c=\dfrac{81}{4}}$

**Solvex ^{2} + 12x + 32 = 0by the method of completing the square**

Solution:

**x ^{2} + 12x + 32 = 0**

=>x

${\left( \dfrac{12}{2}\right) ^{2}=36}$

=>=> x

=>(x + 6)

=>${\sqrt{\left( x+6\right) ^{2}}=\sqrt{4}}$

=>x + 6 = ±2

=> x = (-4, -8)

**Solve 2x ^{2} – 4x – 5 = 0 by the method of completing the square**

Solution:

2x^{2} – 4x – 5 = 0

=>${x^{2}-2x-\dfrac{5}{2}=0}$

=>${x^{2}-2x=-\dfrac{5}{2}}$

${\left( \dfrac{1}{2}\times \left( -2\right) \right) ^{2}=1}$

=>${x^{2}-2x+1=\dfrac{5}{2}+1}$

=>${\left( x-1\right) ^{2}=\dfrac{7}{2}}$

=>${\sqrt{\left( x-1\right) ^{2}}=\pm \sqrt{\dfrac{7}{2}}}$

=>${\sqrt{\left( x-1\right) ^{2}}=\pm \sqrt{\dfrac{7}{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\pm \dfrac{\sqrt{14}}{2}}$

=>${x-1=\pm \dfrac{\sqrt{14}}{2}}$

=>${x=1\pm \dfrac{\sqrt{14}}{2}}$

=>${x=\dfrac{2}{2}\pm \dfrac{\sqrt{14}}{2}}$

=>${x=\left( \dfrac{2+\sqrt{14}}{2},\dfrac{2-\sqrt{14}}{2}\right)}$

Solving a quadratic equation of the form a(x + m)^{2} + n, where a = 1

Expanding (x + m)^{2} + n, we get x^{2} + 2mx + m^{2} + n

Now, if we compare a quadratic equation of the form ax^{2} + bx + c with the above equation, we will obtain the value of m and n

Let us use the equation x^{2} + 12x + 32 = 0

Comparing this equation with the equation x^{2} + 2mx + m^{2} + n, we get

2mx = 12x

=>m = 6

m^{2} + n = 32

=> (6)^{2} + n = 32

=> 36 + n = 32

=> n = -4 Thus, we get the same result (x + 6)^{2 }– 4 as obtained while solving the quadratic equation in solved example 1. The rest of the steps remains the same while solving the quadratic equation, and we will get the result x = (-4, -8)

Last modified on August 3rd, 2023