MULTIPLYING BINOMIALS WITH RADICALS USING ALGEBRAIC
IDENTITIES
Three basic algebraic identities:
(a + b)2 = a2 + 2ab + b2
(a - b)2 = a2 - 2ab + b2
a2 - b2 = (a + b)(a - b)
Some basic operations involving radicals :
After finding the expansion, to do further more simplification, we should know how to simplify radicals.
How to add or subtract radicals ?
We can add or subtract like radicals.
2√3 and 5√3 are like radicals
2√2 and 5√3 are not like radicals
How to multiply radicals ?
To multiply two radicals terms, for example
2√3 x 5√3
Multiply the coefficient of radicals and multiply radicals separately.
2√3 x 5√3 = 10 (3) ==> 30
Expand and simplify :
Problem 1 :
(1 + √2)2
Solution :
(1 + √2)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(1 + √2)2 = 12 + (√2)2 + 2(1) (√2)
= 1 + 2 + 2√2
= 3 + 2√2
So, the answer is 3 + 2√2.
Problem 2 :
(2 - √3)2
Solution :
(2 - √3)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(2 - √3)2 = (2)2 + (√3)2 - 2(2) (√3)
= 4 + 3 - 4√3
= 7 - 4√3
So, the answer is 7 - 4√3.
Problem 3 :
(√3 + 2)2
Solution :
(√3 + 2)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(√3 + 2)2 = (√3)2 + (2)2 + 2(√3) (2)
= 3 + 4 + 4√3
= 7 + 4√3
So, the answer is 7 + 4√3.
Problem 4 :
(1 + √5)2
Solution :
(1 + √5)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(1 + √5)2 = 12 + (√5)2 + 2(1) (√5)
= 1 + 5 + 2√5
= 6 + 2√5
So, the answer is 6 + 2√5.
Problem 5 :
(√2 - √3)2
Solution :
(√2 - √3)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(√2 - √3)2 = (√2)2 + (√3)2 - 2(√2) (√3)
= 2 + 3 - 2√6
= 5 - 2√6
So, the answer is 5 - 2√6.
Problem 6 :
(5 - √2)2
Solution :
(5 - √2)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(5 - √2)2 = (5)2 + (√2)2 - 2(5) (√2)
= 25 + 2 - 10√2
= 27 - 10√2
So, the answer is 27 - 10√2.
Problem 7 :
(√2 + √7)2
Solution :
(√2 + √7)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(√2 + √7)2 = (√2)2 + (√7)2 + 2(√2) (√7)
= 2 + 7 + 2√14
= 9 + 2√14
So, the answer is 9 + 2√14.
Problem 8 :
(4 - √6)2
Solution :
(4 - √6)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(4 - √6)2 = (4)2 + (√6)2 - 2(4) (√6)
= 16 + 6 - 8√6
= 22 - 8√6
So, the answer is 22 - 8√6.
Problem 9 :
(√6 - √2)2
Solution :
(√6 - √2)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(√6 - √2)2 = (√6)2 + (√2)2 - 2(√6) (√2)
= 6 + 2 - 2√12
= 8 - 2√12
So, the answer is 8 - 2√12.
Problem 10 :
(√5 + 2√2)2
Solution :
(√5 + 2√2)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(√5 + 2√2)2 = (√5)2 + (2√2)2 + 2(√5) (2√2)
= 5 + 8 + 4√10
= 13 + 4√10
So, the answer is 13 + 4√10.
Problem 11 :
(√5 - 2√2)2
Solution :
(√5 - 2√2)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(√5 - 2√2)2 = (√5)2 + (2√2)2 - 2(√5) (2√2)
= 5 + 8 - 4√10
= 13 - 4√10
So, the answer is 13 - 4√10.
Problem 12 :
(6 + √8)2
Solution :
(6 + √8)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(6 + √8)2 = (6)2 + (√8)2 + 2(6) (√8)
= 36 + 8 + 12√8
= 44 + 12√8
So, the answer is 44 + 12√8.
Problem 13 :
(5√2 - 1)2
Solution :
(5√2 - 1)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(5√2 - 1)2 = (5√2)2 + (1)2 - 2(5√2) (1)
= 50 + 1 - 10√2
= 51 - 10√2
So, the answer is 51 - 10√2.
Problem 14 :
(3 - 2√2)2
Solution :
(3 - 2√2)2
By using algebraic identity.
(a - b)2 = a2 + b2 - 2ab
We get,
(3 - 2√2)2 = (3)2 + (2√2)2 - 2(3) (2√2)
= 9 + 8 - 12√2
= 17 - 12√2
So, the answer is 17 - 12√2.
Problem 15 :
(1 + 3√2)2
Solution :
(1 + 3√2)2
By using algebraic identity.
(a + b)2 = a2 + b2 + 2ab
We get,
(1 + 3√2)2 = 12 + (3√2)2 + 2(1) (3√2)
= 1 + 18 + 6√2
= 19 + 6√2
So, the answer is 19 + 6√2.
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