Question

Hell guys..solve this.
$\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3} .$
*Find* the value of a and b..​

Answer 1

Given:

$\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3}$

Let's rationalize the given number:

$\frac{(5 + \sqrt{3}) \times (7 +4 \sqrt{3}) }{(7 - 4 \sqrt{3}) \times (7 + 4 \sqrt{3} ) } \\ \\ = \frac{35 + 20 \sqrt{3} + 7 \sqrt{3} + 12 }{ ( {7})^{2} - (4 { \sqrt{3}) }^{2} } \\ \\ = \frac{47 + 27 \sqrt{3} }{49 - 48} = 47 + 27 \sqrt{3}$

47 + 27√3 = a + b√3.

a = 47.

b = 27.

Therefore, the values of a and b are 47 and 27 respectively.

Terms used:

• Rationalisation: The conversion of a irrational number whose denominator has a irrational numbers into a real number where it's denominator is rational.

Answer 2

Step-by-step explanation:

Given:

\frac{5 + \sqrt{3} }{7 - 4 \sqrt{3} } = a + b \sqrt{3}

7−4

3

5+

3

=a+b

3

Let's rationalize the given number:

\begin{gathered}\frac{(5 + \sqrt{3}) \times (7 +4 \sqrt{3}) }{(7 - 4 \sqrt{3}) \times (7 + 4 \sqrt{3} ) } \\ \\ = \frac{35 + 20 \sqrt{3} + 7 \sqrt{3} + 12 }{ ( {7})^{2} - (4 { \sqrt{3}) }^{2} } \\ \\ = \frac{47 + 27 \sqrt{3} }{49 - 48} = 47 + 27 \sqrt{3}\end{gathered}

(7−4

3

)×(7+4

3

)

(5+

3

)×(7+4

3

)

=

(7)

2

−(4

3

)

2

35+20

3

+7

3

+12

=

49−48

47+27

3

=47+27

3

47 + 27√3 = a + b√3.

a = 47.

b = 27.

Therefore, the values of a and b are 47 and 27 respectively.

Terms used: