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)
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)
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?
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)