Question

find the value of 'a' and 'b'

$\frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} } = a + b \sqrt{2}$
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Answer 1

$\huge{\color{lime}{\underbrace{\textsf{\textbf{\color{indigo}{Question:-}}}}}}$

Find the value of a and b :-

$\frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} } = a + b \sqrt{2}$

$\huge{\color{lime}{\underbrace{\textsf{\textbf{\color{indigo}{Answer:-}}}}}}$

$= = > \frac{5 + 3 \sqrt{2} }{5 - 3 \sqrt{2} } \times \frac{5 + 3 \sqrt{2} }{5 + 3 \sqrt{2} }$

$==>$ Rationalising the denominator using the identity a² + b² + 2ab = ( a+b )² and ( a-b ) ( a+b ) = a² - b²:-

$= = > \frac{5²+(3 \sqrt{2)²}+2(5)(3 \sqrt{2)} }{(5)²-(3 \sqrt{2)²} }$

$= = > \frac{25 + 18 + 30 \sqrt{2} }{25 - 18}$

$= = > \frac{43 + 30 \sqrt{2} }{7}$

$==>$ Therefore, a = $\frac{43}{7}$ and b = $\frac{30}{7}$.

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Hope it helps you dear :)

Answer 2

Given :

$:\implies\sf{ \dfrac{5 + 3 \sqrt{2}}{5 - 3 \sqrt{2}} = a + b \sqrt{2}}$

Find :

• Find the value of " a " and " b "

Solution :

$:\implies \sf{\dfrac{5 + 3 \sqrt{2}}{5 - 3 \sqrt{2}} = a + b \sqrt{2}}$

Let's solve the L.H.S side first :

$\sf{LHS = \dfrac{5 + 3 \sqrt{2}}{5 - 3 \sqrt{2}}}$

Rationalising with denominator.

Rationalising refers to multiplying $\sf{(5 + 3 \sqrt{2})}$ with denominator and numerator.

So, then we will get the equation be like :

$:\implies\sf{\dfrac{(5 + 3 \sqrt{2})}{(5 - 3 \sqrt{2})} \times \dfrac{(5 + 3 \sqrt{2})}{(5 + 3 \sqrt{2})}}$

Solving further , making identity :

$:\implies \sf{\dfrac{(5 + 3 \sqrt{2})^{2}}{(5)^{2} - (3 \sqrt{2})^{2}}}$

Identity Using :

$\boxed{\bf{(a + b)^{2} = a^{2} + b^{2} + 2ab}}$

$\boxed{\bf{(a + b) (a - b) = a^{2} - b^{2}}}$

Now,

$:\implies \sf{\dfrac{(5)^{2} + (3 \sqrt{2})^{2} + 2 \times 5 \times (3 \sqrt{2})}{(5)^{2} - (3 \sqrt{2})^{2}}}$

$:\implies \sf{\dfrac{25 + 18 + 10 \times (3 \sqrt{2})}{(25 - (18)}}$

$:\implies \sf{\dfrac{25 + 18 + 30 \sqrt{2})}{7}}$

$:\implies \sf{\dfrac{43 + 30 \sqrt{2})}{7}}$

Now , Comparing it with RHS $\sf{ a + b \sqrt{2}}$

Therefore, from this we get :

$:\implies \bf{\dfrac{43}{7} = a, \dfrac{30}{7} = b}$

$\boxed{\therefore \sf{a = \dfrac{43}{7} \:and\: b = \dfrac{30}{7} = b}}$