**Completing the Square :**

Not every quadratic expression is a perfect squaree trinomial. Completing the square is the process of finding the constant to add to

x^{2} + bx

to create a perfect square trinomial.

The model below depicts the process of completing the square.

To create a perfect square trinomial, add (b/2)^{2} to the variable expression.

x^{2} + bx + (b/2)^{2} = (x + b/2)^{2}

**Example : **

Write x^{2} + 6x + 7 = 0 in the form (x + p)^{2} = q.

**Solution : **

Write the original equation.

x^{2} + 6x + 7 = 0

Isolate the variable expression.

x^{2} + 6x = -7 -----(1)

Determine the constant needed to complete the square.

Comparing x^{2} + bx and x^{2} + 6x, we get

b = 6

So, (b/2)^{2 }= (6/2)^{2} = 3^{2} = 9.

In (1), we have to add 9 to each side.

x^{2} + 6x + 9 = -7 + 9

Write the left side of the equation as a perfect square.

(x + 3)^{2} = 2

Hence, the equation x^{2} + 6x + 7 = 0 can be written as

(x + 3)^{2} = 2

**Example : **

Solve the following quadratic equation using square root :

x^{2} + 12x + 36 = 49

**Solution : **

Write the original equation.

x^{2} + 12x + 36 = 49

Recognize that the quadratic equation is a perfect square trinomial.

x^{2} + 2(6)(x) + 6^{2} = 49

Factor the perfect square trinomial.

(x + 6)^{2} = 49

Take the square root on each side of the equation.

√(x + 6)^{2} = ± √49

x + 6 = ± 7

x + 6 = - 7 or x + 6 = 7

x = - 13 or x = 1

Hence, the solutions are -13 and 1.

**Example : **

Solve the following quadratic equation by completing the square :

x^{2} - 8x - 9 = 0

**Solution : **

Write the original equation.

x^{2} - 8x - 9 = 0

Isolate the variable expression.

x^{2} - 8x = 9 -----(1)

Determine the constant needed to complete the square.

Comparing x^{2} + bx and x^{2} - 8x, we get

b = -8

So, (b/2)^{2 }= (-8/2)^{2} = (-4)^{2} = 16.

In (1), we have to add 16 to each side.

x^{2} - 8x + 16 = 9 + 16

x^{2} - 8x + 16 = 25

Write the left side of the equation as a perfect square.

(x - 4)^{2} = 25

Take the square root on each side of the equation.

√(x - 4)^{2} = ± √25

x - 4 = ± 5

x - 4 = - 5 or x - 4 = 5

x = - 1 or x = 9

Hence, the solutions are -1 and 9.

**Example : **

Write the following quadratic equation in vertex form and graph it :

y = - x^{2} - 2x + 3

What is the maximum or minimum value of the graph of the equation ?

**Solution : **

Write the original equation.

y = - x^{2} - 2x + 3

Factor out the x^{2} coefficient, -1.

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

y = -1(x^{2} - 2 ⋅ x ⋅ 1 - 3)

y = -1(x^{2} - 2 ⋅ x ⋅ 1 - 3)

y = -1(x^{2} - 2 ⋅ x ⋅ 1 + 1^{2} - 1^{2} - 3)

y = - [(x - 1)^{2 }- 1^{2} - 3]

y = - [(x - 1)^{2 }- 1 - 3]

y = - [(x - 1)^{2 }- 4]

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

Hence, the vertex form of the equation y = -x^{2} - 2x + 3 is

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

**Vertex :**

The vertex of the parabola is (1, 4).

**Graph :**

In the given equation y = -x^{2} - 2x + 3, the sign of x^{2 }is negative.

So, its graph is a parabola that opens downward.

The graph of the given quadratic equation has a maximum of y = 4 at x = 1.

**Example :**

Alex plans to create rectangular shaped garden. He has 340 m of fencing available for the garden's perimeter and wants it to have an area of 6000 m^{2}. What dimensions should Alex use ?

**Solution :**

Let x and y be the length and width of the garden respectively.

**Given :** Perimeter = 340

So, we have

2x + 2y = 340

Divide each side by 2.

x + y = 170

Solve for y.

y = 170 - x

Alex wants the area to be 6000 m^{2}.

Write this as an equation.

A = xy

6000 = x(170 - x)

6000 = 170x - x^{2}

x^{2} - 170x = -6000 -----(1)

Determine the constant needed to complete the square.

Comparing x^{2} + bx and x^{2} - 170x, we get

b = -170

So, (b/2)^{2} = (-170/2)^{2 }= (-85)^{2} = 7225.

In (1), we have to add 7225 to each side.

x^{2} - 170x + 7225 = -6000 + 7225

Write the left side of the equation as a perfect square.

(x - 85)^{2} = 1225

Take the square root on each side of the equation.

√(x - 85)^{2} = ± √1225

x - 85 = ± 35

x - 85 = - 35 or x - 85 = 35

x = 50 or x = 110

When x = 50,

y = 170 - 50

y = 120

When x = 110,

y = 170 - 110

y = 60

In each case, there is 340 m of fencing used.

Likewise, the area is 6000 m^{2}.

Hence, Alex should make two sides of the garden 120 m long and the other two sides 50 m long.

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