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# Completing the Square

Sep 17, 2022

## Key Concepts

• Completing Square – Perfect Square Trinomial.
• Understand Vertex Form.

### Introduction

In the previous session, we have learned about solving quadratic equations using square roots and converting the equation into the form x2=d, and taking square root on both sides to find the solution.

Now we will learn about completing the square.

## Trinomial

• What is a trinomial?

A trinomial is a polynomial that consists of three terms.

Example:

1. ax2+bx+c,
1. x+y+z.
• How can you make a quadratic equation into a perfect square trinomial?

Let us see some examples.

### Perfect Square Trinomial

Example 1:

Find the value of ‘c’ that makes the expression x2+4x+c a perfect square trinomial.

Solution:

Given expression x2+4x+c

Let us use the algebraic tiles method to solve this.

The algebraic tiles for x2+4x are:

Now we re-arrange the tiles to make it a square x2+ 4x.

We need to find the value of ‘c’ such that it completes the square.

Since 4 / 2 =2 and 22 = 4

There are four tiles.

Now we convert x2+4x+4 into binomial squared (x+2)2

So, here the ‘c’ value is 4, which makes the expression a perfect square trinomial.

The process of making (b / 2)2 as ‘c’ value for the expression ax2 +bx+c is called completing square.

Example 2:

Find the value of ‘c’ that makes the expression x2+12x+c a perfect square trinomial.

Solution:

Given expression x2+12x+c

Let us use the algebraic tiles method to solve this.

The algebraic tiles for x2+12x are:

Now we re-arrange the tiles to make it a square for x2+12x.

We need to find the value of ‘c’ such that it completes the square

Since 12 / 2 =6 and 62 =36

There are four tiles.

Now we convert x2+12x+36 into binomial squared (x+2)2

So, here the ‘c’ value is 4, which makes the expression a perfect square trinomial.

Example 3:

Find the solutions for the equation x2−10x+15 = 0.

Solution:

Given equation x2−10x+15=0,

Rewrite the given equation in the form ax2+bx = d

We get,  x2−10x = −15

Add 25 on both the sides, as (10 / 2)2 = 25,  we get

10x+25 = -15+25

x²-10x+25 = 10

(x-5)² = 10

Take square root on both the sides, we get

√(x-5)² = ±√/10

(x-5) = ±√/10

Add 5 on both the sides, we get

x-5+5=5± √10

x= 5±√/10

The solutions are 5+ √10 and 5-√10

Example 4:

Find the solutions for the equation x2+8x-9 = 0

Solution:

Given equation x2+8x-9 = 0,

Rewrite the given equation in the form ax2+bx=d

We get,

x2+8x = 9

Adding 16 on both the sides, as (8 / 2)2 = 16,

we get, x²+8x + 16 = 9+16

x²+8x+16= 25

(x+4)² = = 25

Take square root on both the sides, we get

√(x+4)² = ±√/25

(x+4) = ±√/25

Subtracting 4 on both the sides, we get

x+4-4=±√25 +4

x = ±5+4=-1, 9

The solutions are -1 and 9.

### Vertex Form

An equation y = a(x−h)2−k is called the vertex form of quadratic equation, where (h, k) is the vertex.

Vertex form is used in completing square a(x−h)2

Example 1:

Find the solutions for the equation y = x2+2x−6  using vertex form.

Solution:

Given the quadratic equation, y = x2+2x−6

Now we isolate constants to the other side,

y+6 = x²+2x

Adding 1 on both sides, we get

y+6+1 = x²+2x+1

y+7= (x + 1)²

y = (x+1)² -7

The vertex form of the quadratic equation is y = (x+1)2 -7.

Example 2:

Find the solutions for the equation y = 2 +14+25 using vertex form

Solution:

Given the quadratic equation, y = x² + 14x+25

Now we isolate constants to the other side,

y-25 = x² + 14x

Adding 49 on both sides, we get

y-25+49 = x² + 14x + 49

y-25+49 (+7)² y = (x+7)2-24

The vertex form of the quadratic equation is y = (+7)² – 24.

### Vertex Form When a≠1

An equation y = a(x−h)2−k is called Vertex form of quadratic equation, where (h, k) is vertex.

Vertex form is used in completing square a(x−h)2

We have seen the cases when a = 1, now we will see the case when a≠1

Example1:

Find the solutions for the equation y = 5x2+10x+1.

Solution:

y = 5x²+10x+1

y-1 = 5x²+10x

y-1 = 5(x²+2x)

Adding (2/2)2 on both the sides,

y-1+5(1) = 5(x² + 2x+1)

y+4 = 5(x+1)²

y = 5(x+1)² – 4

### Real-Life Example

In a football match, the equation of the kick is recorded as 𝒚 = 𝟒𝒙𝟐+𝟏𝟔𝒙+𝟓. How far the ball went after the kick.

Given equation, y = 4x²+16x+5

y-5 = 4x²+16x

y-5 = 4(x² + 4x)

Adding (4/2)2 on both the sides,

y−5+4(4) = 4(x2+4x+4)

y+11 = 4(x+2)2

The height of the foot after the kick is 11 units and the football reaches the distance of 2 units from the kick point.

## Exercise

1. The coordinates of the vertex of the parabola, whose equation is y = 2x² + 4x – 5 are: a. (2,11) b. (-1,-7) c. (1,1) d. (-2,-5).
2. Find the value of ‘c’ that makes the expression x² + 8x + c a perfect square trinomial.
3. Find the solutions for the equation x² – 10x + 12 = 0.
4. Find the solutions for the equation x²+6x-9= 0
5. Find the solutions for the equation y = x²+4x-6 using vertex form.
6. Find the solutions for the equation y = x²+16x+25 using vertex form.
7. Find the solutions for the equation y = 6x2 + 24x + 2.
8. In a football match, the equation of the kick is recorded as y = 5x²+10x + 5. How far the ball went after the kick.
9. If the average of the roots is 3 and the difference is 2, find the quadratic equation.
10. The coordinates of the vertex of the parabola whose equation is y = x + 4x – 2 are:
1. (2,11)
2. (-1,-7)
3. (1,1)
4. (-2, -6)

### What have we learned

• Solving quadratic equations using completing square method.
• Perfect Square Trinomial.
• Vertex Form.

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