Question

Find the value of a and b if 5+√3\7-4√3=94a+3√3b

Answer 1

Solution :

Now, $\frac{5+\sqrt{3}}{7-4\sqrt{3}}$

  To rationalise the denominator, we multiply both the numerator and the denominator by the conjugate irrational number $(7+4\sqrt{3})$

$=\frac{(5 + \sqrt{3})(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})}$

$=\frac{35+20\sqrt{3}+7\sqrt{3}+12}{49-48}$

$=47+27\sqrt{3}$

$\implies 47+27\sqrt{3}=94a+3\sqrt{3}b$

Comparing between coefficients from both sides, we get

    47 = 94a and $27\sqrt{3} = 3\sqrt{3}b$

⇒ a = $\frac{1}{2}$ and b = 9

Answer 2

Step-by-step explanation:

Now, \frac{5+\sqrt{3}}{7-4\sqrt{3}}

7−4

3

5+

3

To rationalise the denominator, we multiply both the numerator and the denominator by the conjugate irrational number (7+4\sqrt{3})(7+4

3

)

=\frac{(5 + \sqrt{3})(7+4\sqrt{3})}{(7-4\sqrt{3})(7+4\sqrt{3})}=

(7−4

3

)(7+4

3

)

(5+

3

)(7+4

3

)

=\frac{35+20\sqrt{3}+7\sqrt{3}+12}{49-48}=

49−48

35+20

3

+7

3

+12

=47+27\sqrt{3}=47+27

3

\implies 47+27\sqrt{3}=94a+3\sqrt{3}b⟹47+27

3

=94a+3

3

b

Comparing between coefficients from both sides, we get

47 = 94a and 27\sqrt{3} = 3\sqrt{3}b27

3

=3

3

b

⇒ a

2

1

and b = 9