Question

if root2-1/root2+1=a+b root2 then find the value of a and b

Answer 1

Hey friend,
Here is the answer you were looking for:
$\frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} a + b \sqrt{2} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1 } \\ \\ using \: the \: identity \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{2}) }^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1 }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } \\ \\ = \frac{2 +1 - 2 \sqrt{2} }{2 - 1} \\ \\ 3 - 2 \sqrt{2} = a + b \sqrt{2} \\ \\ a = 3 \\ \\ b = - 2$


Hope this helps!!!

@Mahak24

Thanks...
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Answer 2

Step-by-step explanation:

Hey friend,

Here is the answer you were looking for:

\begin{gathered}\frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} a + b \sqrt{2} \\ \\ on \: rationalizing \: the \: denominator \: we \: get \\ \\ = \frac{ \sqrt{2} - 1}{ \sqrt{2} + 1} \times \frac{ \sqrt{2} - 1}{ \sqrt{2} - 1 } \\ \\ using \: the \: identity \\ {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab \\ (a + b)(a - b) = {a}^{2} - {b}^{2} \\ \\ = \frac{ {( \sqrt{2}) }^{2} + {(1)}^{2} - 2 \times \sqrt{2} \times 1 }{ {( \sqrt{2}) }^{2} - {(1)}^{2} } \\ \\ = \frac{2 +1 - 2 \sqrt{2} }{2 - 1} \\ \\ 3 - 2 \sqrt{2} = a + b \sqrt{2} \\ \\ a = 3 \\ \\ b = - 2\end{gathered}

2

+1

2

−1

a+b

2

onrationalizingthedenominatorweget

=

2

+1

2

−1

×

2

−1

2

−1

usingtheidentity

(a−b)

2

=a

2

+b

2

−2ab

(a+b)(a−b)=a

2

−b

2

=

(

2

)

2

−(1)

2

(

2

)

2

+(1)

2

−2×

2

×1

=

2−1

2+1−2

2

3−2

2

=a+b

2

a=3

b=−2