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Simplify the radical expression : (-1-√2)/(1+√2), (√8)/(√2+3) , (√3)/(√27-√3) , (√2-√8)/2(√2-1) , (3√3-2√2)/(√3-√2), (√2-√8)/(2√2-1), (1-√2)/(1+√2), (3√3-2√2)/(√3-√2)

 Nov 20, 2014

Best Answer 

 #2
avatar+26387 
+5

Simplify the radical expression :

1.)  (-1-√2)/(1+√2)

$$\frac{(-1- \sqrt{2} )}{ (1+ \sqrt{2} ) }
=-\frac{(1+ \sqrt{2} )}{ (1+ \sqrt{2} ) } = -1$$

2.)  (√8)/(√2+3)

$$\frac { \sqrt{8} } { \sqrt{2} + 3} = \frac { 2\sqrt{2} } { \sqrt{2} + 3} *\frac{ \sqrt{2}-3 }{ \sqrt{2}-3 } =-\frac{2}{7}(2-3\sqrt{2})$$

3.)  (√3)/(√27-√3)

$$\frac{ \sqrt{3} }{ \sqrt{27}-\sqrt{3} } = \frac{ \sqrt{3} }{ 3\sqrt{3}-\sqrt{3} } = \frac{ \sqrt{3} }{ 2\sqrt{3} } = \frac{1}{2}$$

4.)  (√2-√8)/2(√2-1)

$$\frac{ \sqrt{2}-\sqrt{8} }{ 2(\sqrt{2}-1) } =\frac{ \sqrt{2}-2\sqrt{2} }{ 2(\sqrt{2}-1) } =\frac{ -\sqrt{2} }{ 2( \sqrt{2}-1) } * \frac{ \sqrt{2}+1 }{ \sqrt{2}+1 } = \frac{ -\sqrt{2} } {2} ( \sqrt{2} + 1) = - ( 1 + \frac{ \sqrt{2} } {2} )$$

5.)  (3√3-2√2)/(√3-√2)

$$\frac{ 3\sqrt{3}-2\sqrt{2} } { \sqrt{3}-\sqrt{2} } * \frac{\sqrt{3}+\sqrt{2}}{ \sqrt{3}+\sqrt{2} } = ( 3\sqrt{3}-2\sqrt{2} ) * ( \sqrt{3}+\sqrt{2} ) = 9 + \sqrt{6} - 4 = 5 + \sqrt{6}$$

6.)  (√2-√8)/(2√2-1)

$$\samll{\text{
$
\frac{ \sqrt{2}-\sqrt{8} }{ 2\sqrt{2}-1 }
=\frac{ \sqrt{2}-2\sqrt{2} }{ 2\sqrt{2}-1 }
=\frac{ -\sqrt{2} }{ 2\sqrt{2}-1 } * \frac{ 2\sqrt{2}+1 }{ 2\sqrt{2}+1 }
= \frac{ -\sqrt{2} } {7} ( 2\sqrt{2} + 1)
= -\frac{ 1 } {7} ( 4 + \sqrt{2} )
$
}}$$

7.)  (1-√2)/(1+√2)

$$\small{\text{
$
\frac{ 1-\sqrt{2} } { 1+\sqrt{2} } *\frac{ 1-\sqrt{2} }{ 1-\sqrt{2} }
= -( 1-\sqrt{2} )( 1-\sqrt{2} ) = -(1-2\sqrt{2}+2) = -(3-2\sqrt{2}) = 2\sqrt{2}-3
$
}}$$

again 5.) (3√3-2√2)/(√3-√2)

 Nov 20, 2014
 #1
avatar+23252 
+5

To simplify these, multiply both the numerator and denominator of the fraction by the conjugate of the denominator.

The conjugate of   x + √y  is x - √y.

The conjugate of   x - √y  is x + √y.

The conjugate of  x + a√y  is x - a√y

As an example:

To simplify  (2 + √3) / (4 - √5)  multiply both the numerator and denominator by 4 + √5:

     (2 + √3) / (4 - √5) x (4 + √5) / (4 + √5)

Numerator:  (2 + √3)(4 + √5)  =  8 + 4√3 +2√5 + √15

Denominator:  (4 - √5)(4 + √5)  =  16 - 4√5 + 4√5 - √25  =  16 - 5  = 11

Answer:  ( 8 + 4√3 +2√5 + √15 ) / 11

Any questions about this example?

 Nov 20, 2014
 #2
avatar+26387 
+5
Best Answer

Simplify the radical expression :

1.)  (-1-√2)/(1+√2)

$$\frac{(-1- \sqrt{2} )}{ (1+ \sqrt{2} ) }
=-\frac{(1+ \sqrt{2} )}{ (1+ \sqrt{2} ) } = -1$$

2.)  (√8)/(√2+3)

$$\frac { \sqrt{8} } { \sqrt{2} + 3} = \frac { 2\sqrt{2} } { \sqrt{2} + 3} *\frac{ \sqrt{2}-3 }{ \sqrt{2}-3 } =-\frac{2}{7}(2-3\sqrt{2})$$

3.)  (√3)/(√27-√3)

$$\frac{ \sqrt{3} }{ \sqrt{27}-\sqrt{3} } = \frac{ \sqrt{3} }{ 3\sqrt{3}-\sqrt{3} } = \frac{ \sqrt{3} }{ 2\sqrt{3} } = \frac{1}{2}$$

4.)  (√2-√8)/2(√2-1)

$$\frac{ \sqrt{2}-\sqrt{8} }{ 2(\sqrt{2}-1) } =\frac{ \sqrt{2}-2\sqrt{2} }{ 2(\sqrt{2}-1) } =\frac{ -\sqrt{2} }{ 2( \sqrt{2}-1) } * \frac{ \sqrt{2}+1 }{ \sqrt{2}+1 } = \frac{ -\sqrt{2} } {2} ( \sqrt{2} + 1) = - ( 1 + \frac{ \sqrt{2} } {2} )$$

5.)  (3√3-2√2)/(√3-√2)

$$\frac{ 3\sqrt{3}-2\sqrt{2} } { \sqrt{3}-\sqrt{2} } * \frac{\sqrt{3}+\sqrt{2}}{ \sqrt{3}+\sqrt{2} } = ( 3\sqrt{3}-2\sqrt{2} ) * ( \sqrt{3}+\sqrt{2} ) = 9 + \sqrt{6} - 4 = 5 + \sqrt{6}$$

6.)  (√2-√8)/(2√2-1)

$$\samll{\text{
$
\frac{ \sqrt{2}-\sqrt{8} }{ 2\sqrt{2}-1 }
=\frac{ \sqrt{2}-2\sqrt{2} }{ 2\sqrt{2}-1 }
=\frac{ -\sqrt{2} }{ 2\sqrt{2}-1 } * \frac{ 2\sqrt{2}+1 }{ 2\sqrt{2}+1 }
= \frac{ -\sqrt{2} } {7} ( 2\sqrt{2} + 1)
= -\frac{ 1 } {7} ( 4 + \sqrt{2} )
$
}}$$

7.)  (1-√2)/(1+√2)

$$\small{\text{
$
\frac{ 1-\sqrt{2} } { 1+\sqrt{2} } *\frac{ 1-\sqrt{2} }{ 1-\sqrt{2} }
= -( 1-\sqrt{2} )( 1-\sqrt{2} ) = -(1-2\sqrt{2}+2) = -(3-2\sqrt{2}) = 2\sqrt{2}-3
$
}}$$

again 5.) (3√3-2√2)/(√3-√2)

heureka Nov 20, 2014

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