## Key Concepts

- Analyze Quadratic equations.
- 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:

- ax
^{2}+bx+c,

- 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 x^{2}+4x+c a perfect square trinomial.

Solution:

Given expression x^{2}+4x+c

Let us use the algebraic tiles method to solve this.

The algebraic tiles for x^{2}+4x are:

Now we re-arrange the tiles to make it a square x^{2}+ 4x.

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

Since 4 / 2 =2 and 2^{2} = 4

There are four tiles.

Now we convert x^{2}+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 ax^{2} +bx+c is called completing square.

**Example 2:**

Find the value of ‘c’ that makes the expression x^{2}+12x+c a perfect square trinomial.

Solution:

Given expression x^{2}+12x+c

Let us use the algebraic tiles method to solve this.

The algebraic tiles for x^{2}+12x are:

Now we re-arrange the tiles to make it a square for x^{2}+12x.

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

Since 12 / 2 =6 and 6^{2} =36

There are four tiles.

Now we convert x^{2}+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 x^{2}−10x+15 = 0.

Solution:

Given equation x^{2}−10x+15=0,

Rewrite the given equation in the form ax^{2}+bx = d

We get, x^{2}−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 x^{2}+8x-9 = 0

Solution:

Given equation x^{2}+8x-9 = 0,

Rewrite the given equation in the form ax^{2}+bx=d

We get,

x^{2}+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 = x^{2}+2x−6 using vertex form.

Solution:

Given the quadratic equation, y = x^{2}+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 = 5x^{2}+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 = 𝟒𝒙^{𝟐}+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(x^{2}+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

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

### Concept Map

### What have we learned

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

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