Question

root 2+3/root 2-1=a+b root 2 then b-a is

Answer 1

Answer:

$\frac{\sqrt{2}+3}{\sqrt{2}-1}=a + b \sqrt{2}$

$\frac{\sqrt{2}+3}{\sqrt{2}-1} \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1} = a + b \sqrt{2}$

$\frac{2+\sqrt{2}+3\sqrt{2} + 3}{{(\sqrt{2}) }^{2}-{(1)}^{2}} = a + b \sqrt{2}$

$\frac{5 + 4 \sqrt{2} }{2 - 1} = a + b \sqrt{2}$

$\frac{5 + 4 \sqrt{2} }{1} = a + b \sqrt{2}$

$5 + 4 \sqrt{2} = a + b \sqrt{2}$

$a = 5$

$b \sqrt{2} = 4 \sqrt{2}$

$b = 4$

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Answer 2

Given,

$\bigstar \: \: \boxed{ \large \bf \frac{ \sqrt{2} + 3 }{ \sqrt{2} - 1} = a + b \sqrt{2}}$

• To find : The value of a & b

$<marquee>SOLUTION⤵⤵</marquee>$

Solving LHS, {Rationalising the term of LHS}

$\huge \bf{\dfrac{ \sqrt{2} + 3}{ \sqrt{2} - 1 } \times \dfrac{ \sqrt{2} + 1}{ \sqrt{2} + 1 }} \\ \\ \\ \sf \Longrightarrow \frac{ \sqrt{2} + 3( \sqrt{2} + 1 )}{ {( \sqrt{2} )}^{2} - {1}^{2} } \\ \\ \sf \Longrightarrow \frac{ \sqrt{2} ( \sqrt{2} + 1) + 3( \sqrt{2} + 1)}{2 - 1} \\ \\ \sf \Longrightarrow \frac{2 + \sqrt{2} + 3 \sqrt{2} + 3 }{1} \\ \sf \Longrightarrow 2 + 3 + (1 + 3) \sqrt{2} \\ \sf \Longrightarrow \red5 + \red4 \sqrt{ \red2}$

Now, Equating the obtained value of LHS with RHS,

$\sf5 + 4 \sqrt{2} = a + b \sqrt{2}$

Thus, From above observation, It is concluded that value of a is 5 and b is 4.

Hence,

$\widehat{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

$\boxed{ \tt\Huge a = \red5} \\ \boxed{ \tt\Huge b= \red4}$

$\underbrace{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

• MARK BRAINLIEST ⤵⤵