PRACTICE PROBLEMS ON COMPLETING THE SQUARE

If possible, solve for x using ‘completing the square’ :

(1)  x² + 2x  =  5

(2)  x² - 2x  =  - 7

(3)  x² - 6x  =  - 3

(4)  x² + 10x  =  1

(5)  x² - 4x + 1  =  0

(6)  x² - 2x - 2  =  0

(7)  x² + 2x - 1  =  0

(8)  x² + 2x + 4  =  0

(9)  x² + 3x + 2  =  0

(10)  x²  =  4x + 8

Problem 1 :

x² + 2x  =  5

Solution :

The coefficient of x² is 1, so we don't have to divide the entire equation.

x² + 2x  =  5

Half of the coefficient of x(2/2) is 1.

x² + 2x + 1²  =  5 + 1²

x² + 2⋅x⋅1 + 1²   ==>  a² + 2⋅a⋅b + b²

x² + 2x + 1  =  6

(x + 1)²  =  6

Taking square roots on both sides, we get

(x + 1)  =  ±√6

So, solutions are √6-1 and -√6-1.

Problem 2 :

x² - 2x  =  - 7

Solution :

x² - 2x  =  - 7

Coefficient of x² is 1. So dividing the entire equation is not required.

Coefficient of x is -2, half of it  =  -1

x² - 2x + (-1)²  =  - 7 + (-1)²

x² - 2x + 1  =  - 7 + 1

x² - 2x + 1  =  -6

x² - 2⋅x⋅1 + 1²   ==>  a² - 2⋅a⋅b + b²

(x - 1)²  =  - 6

So, the system has no real solution.

Problem 3 :

x² - 6x  =  - 3

Solution :

x² - 6x  =  - 3

Coefficient of x² is 1. So dividing the entire equation is not required.

Coefficient of x is -6, half of it  =  -3

x² - 6x + (-3)²  =  - 3 + (-3)²

x² - 6x + 9  =  - 3 + 9

x² - 6x + 9  =  6

x² - 2⋅x⋅3 + 3²   ==>  a² - 2⋅a⋅b + b²

(x - 3)²  =  6

(x - 3)  =  ±√6

So, the solutions are √6+3 and -√6+3.

Problem 4 :

x² + 10x  =  1

Solution :

x² + 10x  =  1

x² + 10x + 5²  =  1 + 5²

x² + 10x + 25  =  26

x² + 2⋅x⋅5 + 5²   ==>  a² + 2⋅a⋅b + b²

(x + 5)²  =  26

(x + 5)  =  ±√26

Problem 5 :

x² - 4x + 1  =  0

Solution :

x² - 4x + 1  =  0

x² - 4x  =  - 1

x² - 4x + 2²  =  - 1 + 2²

x² - 2x(2) +  2²  =  - 1 + 4

x² - 2⋅x⋅2 + 2²   ==>  a² - 2⋅a⋅b + b²

(x - 2)²  =  3

Taking the square root on both sides, we get

√(x - 2)²  =  √3

x – 2  =  ± √3

x  =  2 ± √3

So, the value of x is 2 ± √3

Problem 6 :

x² - 2x - 2  =  0

Solution :

x² - 2x - 2  =  0

x² - 2x  =  2

Add (b/2)² on both sides, we get

x² - 2x + (- 1)²  =  2 + (-1)²

x² - 2x + 1  =  2 + 1

x² - 2x + 1  =  3

x² - 2⋅x⋅1 + 1²   ==>  a² - 2⋅a⋅b + b²

(x - 1)²  =  3

Taking the square root of both sides, we get

√(x - 1)²  =  √3

x – 1  =  ± √3

x  =  1 ± √3

So, the value of x is 1 ± √3

Problem 7 :

x² + 2x - 1  =  0

Solution :

x² + 2x - 1  =  0

x² + 2x  =  1

x² + 2x + 1²  =  1 + 1²

x² + 2x + 1  =  2

x² + 2⋅x⋅1 + 1²   ==>  a² + 2⋅a⋅b + b²

(x + 1)²  =  2

Taking the square roots on both sides, we get

√(x + 1)²  =  √2

x + 1  =  ± √2

x  =  - 1 ± √2

So, the value of x is - 1 ± √2

Problem 8 :

x² + 2x + 4  =  0

Solution :

x² + 2x + 4  =  0

x² + 2x  =  - 4

x² + 2x + 1²  =  - 4 + 1²

x² + 2x + 1  =  - 3

(x + 1)²  =  - 3

Taking the square roots on both sides, we get

√(x + 1)²  =   √-3

So, there is no real solution for x.

Problem 9 :

x² + 3x + 2  =  0

Solution :

x² + 3x + 2  =  0

x² + 3x  =  - 2

x² + 3x + (3/2)²  =  - 2 + (3/2)²

x² + 3x + 9/4  =  - 2 + 9/4

x² + 3x + 9/4  =  1/4

(x + 3/2)²  =  1/4

√(x + 3/2)²  =  √1/4

x + 3/2  =  ± 1/2

x  =  1/2 – 3/2 or x  =  - 1/2 – 3/2

x  =  - 2/2 or x  =  - 4/2

x  =  - 1 or x  =  - 2

So, the value of x is – 1 or – 2

Problem 10 :

x²  =  4x + 8

Solution :

By using completing square method x² + bx  =  c,

x²  =  4x + 8

x² - 4x  =  8

x² - 4x + 4  =  8 + 4

x² - 4x + 4  =  12

(x - 2)²  =  12

Taking the square root of both sides, we get

√(x - 2)²  =  √12

x - 2  =  ± √12

x  =  2 ± √12

So, the value of x is 2 ± √12.

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