Question

4/7+4√3 Rationalize the Denominator

Answer 1

Heya user, Here is your answer

$\frac{4}{7 + 4 \sqrt{3} } \\ retionalising \: \\ \frac{4}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \frac{4(7 - 4 \sqrt{3)} }{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} } \: \: \: \: a {}^{2} - {b}^{2} = (a + b)(a - b) \\ \frac{4(7 - 4 \sqrt{3} )}{49 - 16 \times 3} \\ \frac{4(7 - 4 \sqrt{3} }{49 - 48} \\ \frac{4(7 - 4 \sqrt{3}) }{1} \\ 4(7 - 4 \sqrt{3} ) \\ 28 - 16 \sqrt{3}$
Hope it will help you

Answer 2

Answer:

Heya user, Here is your answer

\begin{gathered} \frac{4}{7 + 4 \sqrt{3} } \\ retionalising \: \\ \frac{4}{7 + 4 \sqrt{3} } \times \frac{7 - 4 \sqrt{3} }{7 - 4 \sqrt{3} } \\ \frac{4(7 - 4 \sqrt{3)} }{(7) {}^{2} - (4 \sqrt{3} ) {}^{2} } \: \: \: \: a {}^{2} - {b}^{2} = (a + b)(a - b) \\ \frac{4(7 - 4 \sqrt{3} )}{49 - 16 \times 3} \\ \frac{4(7 - 4 \sqrt{3} }{49 - 48} \\ \frac{4(7 - 4 \sqrt{3}) }{1} \\ 4(7 - 4 \sqrt{3} ) \\ 28 - 16 \sqrt{3} \end{gathered}

7+4

3

4

retionalising

7+4

3

4

×

7−4

3

7−4

3

(7)

2

−(4

3

)

2

4(7−4

3)

a

2

−b

2

=(a+b)(a−b)

49−16×3

4(7−4

3

)

49−48

4(7−4

3

1

4(7−4

3

)

4(7−4

3

)

28−16

3