Question

$\frac{5 + 2 \sqrt{3} }{7 + 4 \sqrt{3} } = a + b \sqrt{3}$
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Answer 1

Using Basic laws which we read in classs 9th:-

the solution is as followed

Answer:-

a=11 and b= -6

Solution!

a) Firstly we rationalised the fraction sum as it is under root in the denominator.

b) Later we tried to solve the Calculations in Numerator part and denominator. simply.

c) we made it to ths form of a+b√3

d) Compared a and b to the given solution.

Atlast,

ignoring my handritting kindly check the attachmemt!!

Answer 2

Answer:

a = 11, b = - 6

Step-by-step explanation:

Given : ${\sf{\ \ {\dfrac{ 5 + 2 {\sqrt{3}} }{ 7 + 4 {\sqrt{3}} }} = a + b {\sqrt{3}} }}$

To Find : a & b

Solution :

Rationalising the denominator, we get

$\Longrightarrow{\sf{ {\dfrac{ 5 + 2 {\sqrt{3}} }{ 7 + 4 {\sqrt{3}} }} \times {\dfrac{ 7 - 4 {\sqrt{3}} }{ 7 - 4 {\sqrt{3}} }} }}$

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${\boxed{\tt{Identity \ : \ (a + b)(a - b) = a^2 - b^2 }}}$

${\tt{Here, \ a = 7, \ b = 4 {\sqrt{3}} }}$

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

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$\Longrightarrow{\sf{ {\dfrac{ (5 + 2 {\sqrt{3}} )( 7 - 4 {\sqrt{3}} )}{ 49 - 48 }} }}$

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$\Longrightarrow{\sf{ {\dfrac{ (5 + 2 {\sqrt{3}} )( 7 - 4 {\sqrt{3}} )}{ 1 }} }}$

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$\Longrightarrow{\sf{ (5 + 2 {\sqrt{3}} )( 7 - 4 {\sqrt{3}} )}}$

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$\Longrightarrow{\sf{ 5 (7 - 4 {\sqrt{3}} ) + 2 {\sqrt{3}} (7 - 4 {\sqrt{3}} ) }}$

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$\Longrightarrow{\sf{ 5(7) + 5(- 4 {\sqrt{3}} ) + 2 {\sqrt{3}} (7) + 2 {\sqrt{3}} (- 4 {\sqrt{3}} ) }}$

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$\Longrightarrow{\sf{ 35 - 20 {\sqrt{3}} + 14 {\sqrt{3}} - 24 }}$

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$\Longrightarrow{\sf{ 11 - 6 {\sqrt{3}} }}$

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We can write this as :

$\Longrightarrow{\sf{ 11 + (- 6) {\sqrt{3}} }}$

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On comparing this with a + b√3, we get

$\Longrightarrow{\boxed{\sf{a = 11}}}$

$\Longrightarrow{\boxed{\sf{b = - 6}}}$