Question

please answer these both questions ​

Answer 1

Answer:

(i) a= 11/7 ,b=6/7

(ii) a=11 ,b=-6

Answer 2

Step-by-step explanation:

(1) $\bold{\dfrac{3+\sqrt{2} }{3-\sqrt{2} }=a+b\sqrt{2}}$

Rationalizing L.H.S :-

$\rightarrow \bold{\dfrac{3+\sqrt{2} }{3-\sqrt{2}\ }\bold{\times\dfrac{3+\sqrt{2} }{3+\sqrt{2} } }}$

$\rightarrow \bold{\dfrac{(3+\sqrt{2})^2 }{(3+\sqrt{2})(3-\sqrt{2}) } }$

We know :-

• (a + b)² = a² + b² + 2ab
• (a² - b²) = (a + b)(a - b)

Using them here :-

$\rightarrow \bold{\dfrac{(3)^2+(\sqrt{2})^2+2(3)(\sqrt{2}) }{(3)^2-(\sqrt{2})^2 } }$

$\rightarrow \bold{\dfrac{9+2+6\sqrt{2} }{9-2} }$

$\rightarrow \bold{\dfrac{11+6\sqrt{2} }{7} }$

$\rightarrow \bold{\dfrac{11}{7}+\dfrac{6\sqrt{2} }{7} }$

∴ $\bold{\dfrac{11}{7}+\dfrac{6\sqrt{2} }{7} }$ = $\bold{a+b\sqrt{2} }$

Comparing them we get,

• $\bold{a=\dfrac{11}{7} }$ and $\bold{b=\dfrac{6}{7} }$.

(2) $\bold{\dfrac{5+2\sqrt{3} }{7+4\sqrt{3} } =a+b\sqrt{3} }$

Rationalizing L.H.S :-

$\rightarrow \bold{\dfrac{5+2\sqrt{3} }{7+4\sqrt{3} } \times\dfrac{7-4\sqrt{3} }{7-4\sqrt{3} } }$

$\rightarrow \bold{\dfrac{5(7-4\sqrt{3})+2\sqrt{3}(7-4\sqrt{3}) }{(7)^2-(4\sqrt{3})^2} }$

$\rightarrow \bold{\dfrac{35-20\sqrt{3}+14\sqrt{3}-24 }{49-48} }$

$\rightarrow \bold{\dfrac{11-6\sqrt{3} }{1} }$

∴ $\bold{\dfrac{11-6\sqrt{3} }{1} =a+b\sqrt{3} }$

Comparing then we get :-

• $\bold{a=11}$ and $\bold{b=-6}$.

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