Question

CLASS 9
I need answers for both of them... atleast the 2nd one...
THESE QUESTIONS ARE BASED ON THE CH (Number System)​

Answer 1

$1.) \: \frac{1}{ \sqrt{3} - \sqrt{2} } \\ = \frac{1}{ \sqrt{3} - \sqrt{2} } \times \frac{ \sqrt{3} + \sqrt{2} }{ \sqrt{3} + \sqrt{2} } \\ = \frac{ \sqrt{3} + \sqrt{2} }{ { (\sqrt{3}) }^{2} - ( { \sqrt{2} )}^{2} } \\ = \frac{1.732 + 1.414}{3 - 2} \\ = \frac{3.146}{1} \\ = 3.146$

Answer 2

Question :

If $\sf{\sqrt{3} = 1.732}$ and $\sf{\sqrt{2} = 1.414}$, then find the value of $\sf{\dfrac{1}{\sqrt{3} - \sqrt{2}}}$

Answer :

• The value of $\sf{\dfrac{1}{\sqrt{3} - \sqrt{2}}}$ is $\sf{\dfrac{1}{1.318}}$

Explanation :

Given :

• The value of $\sf{\sqrt{3} = 1.732}$
• $\sf{\sqrt{2} = 1.414}$

To find :

• The value of $\sf{\dfrac{1}{\sqrt{3} - \sqrt{2}}}$ = ?

Solution :

By substituting the value of √3 and √2 in the equation, we get :

$:\implies \sf{\dfrac{1}{\sqrt{3} - \sqrt{2}} = \dfrac{1}{\sqrt{3} - \sqrt{2}}} \\ \\ \\ :\implies \sf{\dfrac{1}{\sqrt{3} - \sqrt{2}} = \dfrac{1}{1.732 - 1.414}} \\ \\ \\ :\implies \sf{\dfrac{1}{\sqrt{3} - \sqrt{2}} = \dfrac{1}{1.318}} \\ \\ \\ \boxed{\therefore \sf{\dfrac{1}{\sqrt{3} - \sqrt{2}} = \dfrac{1}{1.318}}}$

Therefore,

• The value of $\sf{\dfrac{1}{\sqrt{3} - \sqrt{2}}}$ = $\sf{\dfrac{1}{1.318}}$

Question :

Find the value of a and b for $\sf{a + b\sqrt{3} = \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

Answer :

The value of a and b for the equation $\sf{a + b\sqrt{3} = \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$ is

Explanation :

Given :

• Equation with whose respect , we have to find the value of a and b , a = 7 ; b = 3.

$\sf{a + b\sqrt{3} = \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

To find :

• The value of a and b. a = ? ; b = ?

Solution :

From the given equation, we get :

$\sf{a + b\sqrt{3} = \dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

• L.H.S. = $\sf{a + b\sqrt{3}}$

• R.H.S. = $\sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

Here, let's solve the R.H.S. of the equation.

$:\implies \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}}}$

By multiplying $\sf{(2 + \sqrt{3})}$ to both the Numerator and Denominator of the equation, we get :

$:\implies \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{(2 + \sqrt{3}) \times (2 + \sqrt{3})}{(2 - \sqrt{3}) \times (2 + \sqrt{3})}} \\ \\ \\ :\implies \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{2(2 + \sqrt{3}) + \sqrt{3}(2 + \sqrt{3})}{2(2 + \sqrt{3}) - \sqrt{3}(2 + \sqrt{3})}} \\ \\ \\ :\implies \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{4 + 2\sqrt{3} + 2\sqrt{3} + 3}{4 + 2\sqrt{3} - 2\sqrt{3} - 3}} \\ \\ \\ :\implies \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = \dfrac{7 + 4\sqrt{3}}{1}} \\ \\ \\\boxed{\therefore \sf{\dfrac{2 + \sqrt{3}}{2 - \sqrt{3}} = 7 + 4\sqrt{3}}}$

Now by comparing the L.H.S. and R.H.S. of the equation, we get :

$:\implies \sf{a + b\sqrt{3} = 7 + 3\sqrt{3}} \\ \\ \\ :\implies \sf{a = 7, b\sqrt{3} = 3\sqrt{3}} \\ \\ \\ :\implies \sf{a = 7, b = \dfrac{3\sqrt{3}}{\sqrt{3}}} \\ \\ \\ :\implies \sf{ a = 7, b = 3} \\ \\ \\ \boxed{\therefore \sf{a = 7; b = 3}}$

Hence the value of a is 7 and that of b is 3.