SOLVING QUADRATIC EQUATIONS BY COMPLETING THE SQUARE WORKSHEET

Solve each of the following quadratic equations by completing the square method.

Problem 1 :

9x² - 12x + 4  =  0

Problem 2 :

x² - 6x - 16  =  0

Problem 3 :

x² - 5x + 6  =  0

Problem 4 :

(5x + 7)/(x - 1)  =  3x + 2

Answers

Problem 1 :

9x² - 12x + 4  =  0

Solution :

Step 1 :

In the given quadratic equation 9x² - 12x + 4 = 0, divide the complete equation by 9 (coefficient of x²). 

  x² - (12/9)x + (4/9)  =  0

x² - (4/3)x + (4/9)  =  0

Step 2 :

Subtract 4/9 from each side. 

x² - (4/3)x  =  - 4/9

Step 3 :

In the result of step 2, write the "x" term as a multiple of 2. 

Then, 

x² - (4/3)x  =  - 4/9

x² - 2(x)(2/3)  =  - 4/9

Step 4 :

Now add (2/3)² to each side to complete the square on the left side of the equation.  

Then, 

x² - 2(x)(2/3) + (2/3)²  =  - 4/9 + (2/3)²

(x - 2/3)²  =  - 4/9 + 4/9

(x - 2/3)²  =  0

Take square root on both sides. 

√(x - 2/3)²  =  √0

x - 2/3  =  0

Add 2/3 to each side. 

x  =  2/3

So, the solution is 2/3. 

Problem 2 :

x² - 6x - 16  =  0

Solution :

Step 1 :

In the quadratic equation x² - 6x - 16 = 0, the coefficient of x² is 1. 

So, we have nothing to do in this step. 

Step 2 :

Add 16 to each side of the equation x² - 6x - 16  =  0.

x² - 6x  =  16

Step 3 :

In the result of step 2, write the "x" term as a multiple of 2. 

Then, 

x² - 6x  =  16

x² - 2(x)(3)  =  16

Step 4 :

Now add 3² to each side to complete the square on the left side of the equation.  

Then, 

x² - 2(x)(3) + 3²  =  16 + 3²

(x - 3)²  =  16 + 9

(x - 3)²  =  25

Take square root on both sides. 

√(x - 3)²  =  √25

x - 3  =  ±5

x - 3  =  -5  or  x - 3  =  5

x  =  -2  or  x  =  8

So, the solution is {-2, 8}. 

Problem 3 :

x² - 5x + 6  =  0

Solution :

Step 1 :

In the quadratic equation x² - 5x + 6 = 0, the coefficient of x² is 1. 

So, we have nothing to do in this step. 

Step 2 :

Subtract 6 from each side of the equation x² - 5x + 6 = 0.

x² - 5x  =  -6

Step 3 :

In the result of step 2, write the "x" term as a multiple of 2. 

Then, 

x² - 5x  =  -6

x² - 2(x)(5/2)  =  -6

Step 4 :

Now add (5/2)² to each side to complete the square on the left side of the equation.  

Then, 

x² - 2(x)(5/2) + (5/2)²  =  -6 + (5/2)²

(x - 5/2)²  =  -6 + 25/4

(x - 5/2)²  =  -24/4 + 25/4

(x - 5/2)²  =  (-24 + 25)/4

(x - 5/2)²  =  1/4

Take square root on both sides. 

√(x - 5/2)²  =  √(1/4)

x - 5/2  =  ±1/2

x - 5/2  =  -1/2  or  x - 5/2  =  1/2

x  =  -1/2 + 5/2  or  x  =  1/2 + 5/2

x  =  4/2  or  x  =  6/2

x  =  2  or  x  =  3  

So, the solution is {2, 3}. 

Problem 4 :

(5x + 7)/(x - 1)  =  3x + 2

Solution :

Write the given quadratic equation in the form :

ax² + bx + c  =  0

Then, 

(5x + 7)/(x - 1)  =  3x + 2

Multiply each side by (x - 1). 

5x + 7  =  (3x + 2)(x - 1)

Simplify. 

5x + 7  =  3x² - 3x + 2x - 2

5x + 7  =  3x² - x - 2

0  =  3x² - 6x - 9

or

3x² - 6x - 9  =  0

Divide the entire equation by 3.

x² - 2x - 3  =  0

Step 1 :

In the quadratic equation x² - 2x - 3 = 0, the coefficient of x² is 1. 

So, we have nothing to do in this step. 

Step 2 :

Add 3 to each side of the equation x² - 2x - 3 = 0.

x² - 2x  =  3

Step 3 :

In the result of step 2, write the "x" term as a multiple of 2. 

Then, 

x² - 2x  =  3

x² - 2(x)(1)  =  3

Step 4 :

Now add 1² to each side to complete the square on the left side of the equation.  

Then, 

x² - 2(x)(1) + 1²  =  3 + 1²

(x - 1)²  =  3 + 1

(x - 1)²  =  4

Take square root on both sides. 

√(x - 1)²  =  √4

x - 1  =  ±2

x - 1  =  -2  or  x - 1  =  2

x  =  -1  or  x  =  3

So, the solution is {-1, 3}. 

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