Question

SM+ 8
$1 \div 8 + 3 \sqrt{5}$
​

Answer 1

Correct Question:-

$\mathcal{Rationalis \: the \: denominator } \: \: \frac{1}{8 + 3 \sqrt{5} }$

Solution:-

Let's solve the problem

We have,

$\frac{1}{8 + 3 \sqrt{5} }$

The denomination is 8+3√5. Multiplying the numerator and denomination by 8-3√5, We get

$⟹ \frac{1}{8 + 3 \sqrt{5} } \times \frac{8 - 3 \sqrt{5} }{8 - 3 \sqrt{5} }$

$⟹ \frac{8 - 3 \sqrt{5} }{(8 + 3 \sqrt{5})(8 - 3 \sqrt{5} ) }$

⬤ Applying Algebraic Identity

(a+b)(a-b) = a² - b² to the denominator

We get,

$⟹ \frac{8 - 3 \sqrt{5} }{(8 {)}^{2} -( 3 \sqrt{5} {)}^{2} }$

$⟹ \frac{8 - 3 \sqrt{5} }{64 - 45}$

$⟹ \frac{8 - 3 \sqrt{5} }{19}$

Hence, the denominator is rationalised.

Answer:-

$\frac{8 - 3 \sqrt{5} }{19}$

• I hope it's help you...☺

Know more Algebraic Identities:-

(a+ b)² = a² + b² + 2ab

( a - b )² = a² + b² - 2ab

( a + b )² + ( a - b)² = 2a² + 2b²

( a + b )² - ( a - b)² = 4ab

( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ca

a² + b² = ( a + b)² - 2ab

(a + b )³ = a³ + b³ + 3ab ( a + b)

( a - b)³ = a³ - b³ - 3ab ( a - b)

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Answer 2

Correct solution:-

1/(8+3√5)

= 1/(8+3√5) x (8-3√5)/(8-3√5)

= (8-3√5)/(8+3√5)(8-3√5)

= (8-3√5)/(8)²-(3-√5)²

= (8-3√5)/(64-45)

= (8-3√5)/19 is the correct answer.