Question

Find the values of a and h (√2+1/√2-1)-(√2-1/√2+1)=a+√2b

Answer 1

Given,$( \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } ) -(\frac{ \sqrt{2}- 1}{\sqrt{2}+ 1} ) =a +b \sqrt{2}$

To find,

Values of a and b.

Main solution :

Firstly, simplyfying the LHS $(\frac{\sqrt{2}+1}{ \sqrt{2}-1}\times\frac{\sqrt{2} + 1 }{ \sqrt{2}+1} )-( \frac{ \sqrt{2}-1 }{ \sqrt{2} +1 }\times\frac{ \sqrt{2}- 1}{\sqrt{2} - 1 } )$

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

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

$(3 + 2 \sqrt{2} ) - (3 - 2 \sqrt{2} )$

$3 + 3 \sqrt{2}- 3 + 2 \sqrt{2}$

$0+4\sqrt{2}$

Now, comparing this result to RHS:-

$0+4\sqrt{2}= a+b\sqrt{2}$

We can see that:-a = 0 and b = 4

Answer :- a=0;b=4

Answer 2

Hey there !

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Given :

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

To find the value of a and b , we need to simplify LHS , by rationalising it's denominator.

We have;

$\bold{( \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } ) - ( \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } ) = a + \sqrt{2} b} \\ \\ \\ \\ \bold{\frac{ \sqrt{2} + 1}{ \sqrt{2} - 1 } \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1} } - \bold{\frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 } \times \frac{ \sqrt{2} - 1 }{ \sqrt{2} - 1} } \\ \\ \\ \bold{ \frac{( \sqrt{2} + 1) {}^{2} }{( \sqrt{2}) {}^{2} - (1) {}^{2} } - \frac{( \sqrt{2} - 1) {}^{2} }{( \sqrt{2}) {}^{2} - (1) {}^{2}} } \\ \\ \\ \bold{ \frac{2 + 1 + 2 \sqrt{2} }{2 - 1} - \frac{2 + 1 - 2 \sqrt{2} }{2 - 1} } \\ \\ \\ \bold{\cancel 3 + 2 \sqrt{2} - \cancel 3 + 2 \sqrt{2} } \\ \\ \\ \bold{ 0 + 4 \sqrt{2} }$

Now, on comparing LHS with RHS, we get ;

$\bold{0 + 4 \sqrt{2} = a + b \sqrt{2} }$

Therefore, the value of a and b are ;

$\therefore \: {\bold{a = 0 \: \: and \: \: b = 4}}$

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Thanks for the question !