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
x + 1 = √6 x = √6-1 |
x + 1 = -√6 x = -√6-1 |
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
x - 3 = √6 x = √6+3 |
x - 3 = -√6 x = -√6+3 |
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
x + 5 = √26 x = √26-5 |
x + 5 = -√26 x = -√26-5 |
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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