**ID#:** **080600100003**

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**COMPLETING THE SQUARE**

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- A quadratic expression is one of the form aX² + bX + c, where a, b & c are constants.
- We may not always be able to factorize a quadratic expression, thus, in order to simplify it, we write it in the form, a(X + h) ² + k or k- a(X +h) ²
- (X + h) ² = X² + 2hX + h², this expression is a perfect square.

**STRATEGY FOR SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE.**

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- Make the coefficient of X² unity or 1.
- Add half the coefficient of X all squared.
- Because we have added this amount in step 2, we will have to subtract it at the end of the equation.
- We can now solve for X since the equation had been simplified.

**EXAMPLE 1): **X² + 6X + 5 = 0, solve this equation by completing the square.

Step 1): The coefficient of X² is already 1.

Step 2): The coefficient of X is 6, we need to all half of 6 all squared, that is, (3) ².

The equation now becomes: X² + 6X + (3) ² + 5 = 0

Step 3): Because we have added (3) ², we need to subtract it back at the end of the equation. The equation now becomes:

X² + 6X + (3) ² + 5 – (3) ² = 0

- (X + 3) ² + 5 – 9 = 0
- (X + 3) ² – 4 = 0

Step 4): Solving for X, from the above equation,

(X + 3) ² – 4 = 0

- (X + 3) ² = 4
- (X + 3) = ±√ 4
- (X + 3) = ±2
- X = ±2 – 3
**X = 2- 3 = -1 OR X = -2 -3 = -5**

**EXAMPLE 2): **X² – 3X – 6 = 0, solve this equation by completing the square.

Step 1): The coefficient of X² is already 1.

Step 2): The coefficient of X is -3. We need to add half of -3 all squared, that is, (-3/2) ².

The equation now becomes: X² – 3X + (-3/2) ² -6 = 0

Step 3): Because we have added (-3/2) ², we need to subtract it back at the end of the equation. The equation now becomes:

X² – 3X + (-3/2) ² – 6 – (-3/2) ² = 0

=>(X – 3/2) ² – 6 – (9/4) = 0

=> (X – 3/2) ² – 8¼ = 0

Step 4): Solving for X from the above equation:

(X – 3/2) ² – 8¼ = 0

=>(X – 3/2) ² = 8¼

=> (X – 3/2) = ±√8¼

=> X = ±√8¼ + 3/2

=> **X = +****√****8¼ + 3/2 OR X= –****√****8¼ + 3/2**

**EXAMPLE 3): **2X² + 3X – 2 = 0, solve this equation by completing the square.

NOTE: In this equation, the coefficient of X² ≠ 1

Step 1): 2X² + 3X – 2 = 0, becomes, [2X² + 3X] – 2 = 0

We now have to make the coefficient on X², 1. Thus, the equation becomes:

2[X² + 3/2X] -2 = 0

Step 2): The coefficient of X = 3/2.

We are required to add half of 3/2, squared = (3/2 x ½)²

= (¾)²

Therefore, the equation now becomes: 2[X² + 3/2X + (¾)²] -2 = 0

=> 2[X + ¾]² -2 = 0

Step 3): Because we have added (¾)², we now have 2 subtract this back at the end of the

equation, but one must however keep in mind that (¾)² has to be multiplied by 2,

since when it was added, it is being multiplied by 2. The equation now becomes:

2[X + ¾]² -2 – 2(¾)² = 0

=>2[X + ¾]² – 25/8 = 0

Step 4): Solving the equation: 2[X + ¾]² – 25/8 = 0

=>2[X + ¾]² = 25/8

=> [X + ¾]² = 25/8 x ½

=> [X + ¾]² = 25/16

=> [X + ¾] = ±√ 25/16

=> [X + ¾] = ±5/4

=> X = ±5/4 – ¾

=> **X = 5/4 – ¾ = ½ OR X = -5/4 – ¾ = -2**

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**PRACTICE QUESTIONS **

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Express each of the following in the form a(X +h) ² + k OR k-a(X +h) ²

- X² + 6X + 5
- X² – 8X – 3
- 3X² + 4X -5

4) X² – 9X -1

- -2X² +5X +7

Solve the following equations, using the method of completion of squares.

- X² + 9X + 10 = 0
- 4X² + 7X -5 = 0
- 3X² + 2X -2 = 0

Source: Caribbean Secondary Education Certificate Mathematics