HOW TO WRITE AN EXPRESSION AS THE SQUARE OF A BINOMIAL
Expansion of (a + b)2 :
(a + b)2 = (a + b)(a + b)
Use the FOIL method to multiply the two binomials on the right side.
(a + b)2 = a ⋅ a + ab + ab + b ⋅ b
(a + b)2 = a2 + 2ab + b2
The trinomial a2 + 2ab + b2 is a perfect square trinomial. Because it can be written as the square of a binomial, that is (a + b)2.
When a trinomial is in the form a2 + 2ab + b2, it can be written as the square of a binomial.
Expansion of (a - b)2 :
(a + b)2 = (a + b)(a + b)
Use the FOIL method to multiply the two binomials on the right side.
(a - b)2 = a ⋅ a - ab - ab + b ⋅ b
(a - b)2 = a2 - 2ab + b2
The trinomial a2 - 2ab + b2 is also a perfect square trinomial. Because it can be written as the square of a binomial, that is (a + b)2.
When you have a trinomial in the form a2 - 2ab + b2, it can be written as the square of a binomial.
Write each expression in terms of square of a binomial.
Example 1 :
x2 + 6x
Solution :
In x2 + 6x, write 6x as a multiple of 2. That is, in the form of 2ab.
x2 + 6x = x2 + 2(x)(3)
x2 + 2(x)(3) is in the form of a2 + 2ab.
Comparing
x2 + 2(x)(3)
and
a2 + 2ab + b2,
instead b2, we must have +32 in x2 + 2(x)(3). But +32 is not there. So, add 32 and subtract 32 in x2 + 2(x)(3).
x2 + 6x = x2 + 2(x)(3) + 32 - 32
x2 + 2(x)(3) + 32 is in the form of a2 + 2ab + b2. So, you can write x2 + 2(x)(3) + 32 as the square of the binomial. That is (x + 3)2.
Therefore,
x2 + 6x = (x + 3)2 - 32
x2 + 6x = (x + 3)2 - 9
Example 2 :
x2 - 8x
Solution :
In x2 - 8x, write 8x as a multiple of 2. That is, in the form of 2ab.
x2 - 8x = x2 - 2(x)(4)
x2 - 2(x)(4) is in the form of a2 - 2ab.
Comparing
x2 - 2(x)(4)
and
a2 - 2ab + b2,
instead b2, we must have +42 in x2 - 2(x)(4). But +42 is not there. So, add 42 and subtract 42 in x2 - 2(x)(4).
x2 - 8x = x2 - 2(x)(4) + 42 - 42
x2 - 2(x)(4) + 42 is in the form of a2 + 2ab + b2. So, you can write x2 - 2(x)(4) + 42 as the square of the binomial. That is (x - 4)2.
Therefore,
x2 - 8x = (x - 4)2 - 42
x2 - 8x = (x - 3)2 - 16
Example 3 :
x2 + 12x
Solution :
x2 + 12x = x2 + 2(x)(6)
= x2 + 2(x)(6) + 62 - 62
= (x + 6)2 - 62
= (x + 6)2 - 36
Example 4 :
x2 - 3x
Solution :
x2 - 3x = x2 - 2(x)(3/2)
= x2 - 2(x)(3/2) + (3/2)2 - (3/2)2
= (x - 3/2)2 - (3/2)2
= (x - 3/2)2 - 9/4
Example 5 :
2x2 + 20x
Solution :
2x2 + 20x = 2(x2 + 10x)
= 2[x2 + 2(x)(5)]
= 2[x2 + 2(x)(5) + 52 - 52]
= 2[(x + 5)2 - 52]
= 2[(x + 5)2 - 25]
= 2(x + 5)2 - 50
Solving Quadratic Equations by Completing the Square Method
Example 6 :
Solve :
x2 + 6x + 9 = 0
Solution :
x2 + 6x + 9 = 0
x2 + 2(x)(3) + 9 = 0
x2 + 2(x)(3) + 32 = 0
(x + 3)2 = 0
Take square root on both sides.
x + 3 = 0
Subtract 3 from both sides.
x = -3
Example 7 :
Solve :
x2 - 2x - 8 = 0
Solution :
x2 - 2x - 8 = 0
x2 - 2(x)(1) - 8 = 0
x2 - 2(x)(1) + 12 - 12 - 8 = 0
(x2 - 2(x)(1) + 12) - 12 - 8 = 0
(x - 1)2 - 1 - 8 = 0
(x - 1)2 - 9 = 0
Add 9 to both sides.
(x - 1)2 = 9
Take square root on both sides.
x - 1 = ±√9
x - 1 = ±3
x - 1 = 3 or x - 1 = -3
x = 4 or x = -2
Example 8 :
Solve :
3x2 - 12x + 2 = 0
Solution :
3x2 - 12x + 2 = 0
3(x2 - 4x) + 2 = 0
3[x2 - 2(x)(2) + 22 - 22] + 2 = 0
3[(x - 2)2 - 22] + 2 = 0
3[(x - 2)2 - 4] + 2 = 0
3(x - 2)2 - 12 + 2 = 0
3(x - 2)2 - 10 = 0
Add 10 to both sides.
3(x - 2)2 = 10
Divide both sides by 3.
(x - 2)2 = 10/3
Take square root on both sides.
x - 2 = ±√(10/3)
x - 2 = ±√(10/3)
x - 2 = √(10/3) or x - 2 = -√(10/3)
x = 2 + √(10/3) or x = 2 - √(10/3)
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