USIND DISTRIBUTIVE PROPERTY WITH RADICALS

When we use distributive property, the following operations may be useful.

Expand and simplify 

Example 1 :

√2 (√5+√2)

Solution  :

Given, √2 (√5+√2)

By distribution, we get

√2 (√5+√2)  =  √2×√5 + √2×√2

=  √10+2

So, the answer is √10+2.

Example 2 :

√3 (1-√3)

Solution  :

Given, √3 (1-√3)

By distribution, we get

√3 (1-√3)  =  √3×1 - √3×√3

=  √3-3 

Example 3 :

√11 (2√11-1)

Solution  :

Given, √11 (2√11-1)

By distribution, we get

√11 (2√11-1)  =  √11×2√11 - √11×1

=  2×11 - √11

=  22-√11

Example 4 :

2√3 (√3-√5)

Solution  :

Given, 2√3 (√3-√5)

By distribution, we get

2√3 (√3-√5)  =  (2√3×√3)-(2√3×√5)

=  (2x3) - 2√15

=  6-2√15 

Example 5 :

(1+√2) (2+√2)

Solution  :

Given, (1+√2) (2+√2)

By distribution, we get

(1+√2) (2+√2)  =  2+√2+2√2+2

=  4+3√2

Example 6 :

(√3+2) (√3-1)

Solution  :

Given, (√3+2) (√3-1)

By distribution, we get

(√3+2) (√3-1)  =  (√3×√3)-(√3×1)+(2×√3)-(2×1)

=  3-√3+2√3-2

=  1+√3 

Example 7 :

(√5+2) (√5-3)

Solution  :

Given, (√5+2) (√5-3)

By distribution, we get

(√5+2) (√5-3)  =  (√5×√5)-(√5×3)+(2×√5)-(2×3)

=  5-3√5+2√5-6

=  -1-√5

Example 8 :

(2√2+√3) (2√2-√3)

Solution  :

Given, (2√2+√3) (2√2-√3)

By algebraic identity, we get

(2√2+√3) (2√2-√3)  =  (2√2)²-(√3)²

=  4×2 - 3

=  5

Example 9 :

(2+√3) (2+√3)

Solution  :

Given, (2+√3) (2+√3)

By distribution, we get

(2+√3) (2+√3)  =  2×2+(2×√3)+(√3×2)+(√3×√3)

=  4+2√3+2√3+3

=  7+4√3

Example 10 :

(4-√2) (3+√2)

Solution  :

Given, (4-√2) (3+√2)

By distribution, we get

(4-√2) (3+√2)  =  (4×3)+(4×√2)-(√2×3)-(√2×√2)

=  12+4√2-3√2-2

=  10+√2 

Example 11 :

(√7-√3) (√7+√3)

Solution  :

Given, (√7-√3) (√7+√3)

By using algebraic identity

(a+b)(a-b)  =  a²-b²

(√7-√3) (√7+√3)  =  (√7)²-( √3 )²

=  7-3

=  4

Example 12 :

(4-√2) (3-√2)

Solution  :

Given, (4-√2) (3-√2)

By distribution, we get

(4-√2) (3-√2)  =  4×3 - (4×√2) - (√2×3) + (√2×√2)

=  12-4√2-3√2+2

=  14-7√2

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