Question

rationalise the following denominator
1. 3 root 5 + root 3 /root 5 – root 3

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Answer 1

Solution:

The given fraction is represented as:

$\tt{\dfrac{3 \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } }$

While rationalizing the denominator, multiply numerator and denominator by the conjugate of denominator.

Here conjugate of √5 - √3 is √5 + √3

So, multiply the numerator and denominator by √5 + √3

$\dashrightarrow \: \: \tt{ \dfrac{3 \sqrt{5} + \sqrt{3} }{ \sqrt{5} - \sqrt{3} } \times \dfrac{ \sqrt{5} + \sqrt{3} }{ \sqrt{5} + \sqrt{3} } }$

Use the identity (a + b) (a - b) = a² - b² in the denominator

$\dashrightarrow \: \: \tt{ \dfrac{(3 \sqrt{5} + \sqrt{3}) ( \sqrt{5} + \sqrt{3} )}{ (\sqrt{5})^{2} - (\sqrt{3}) ^{2} } }$

$\to \: \sf{ \dfrac{3 \sqrt{5} ( \sqrt{5} + \sqrt{3} ) + \sqrt{3}( \sqrt{5} + \sqrt{3} ) }{5 - 3} }$

On multiplying,

$\dashrightarrow \: \: \tt{ \dfrac{15 + 3 \sqrt{15} + \sqrt{15} + 3}{2} }$

$\dashrightarrow \: \: \tt{ \dfrac{18 + 4 \sqrt{15} }{2} }$

Taking out 2 as a common factor,

$\dashrightarrow \: \: \tt{ \dfrac{ \cancel{2}(9 + 2 \sqrt{15} )}{ \cancel{2}} }$

$\implies \: \: \boxed{ \bf{9 + 2 \sqrt{15} }}$

Answer 2

Step-by-step explanation:

Given:-

(3√5 + √3)/(√5 - √3)

To find out:-

Rationalised values of denominator.

Solution:-

We have

(3√5 + √3)/(√5 - √3)

The denominator = √5 - √3

We know that

Rationalising factor of √a + √b = √a - √b

So, the rationalising factor of √5 - √3 = √b + √3

On Rationalising the denominator them

→ [(3√5+√3)/(√5-√3)]×[(√5+√3)/(√5+√3)]

→ [(3√5+√3)(√5+√3)]/[(√5-√3)(√5+√3)]

Conider (√5-√3)(√5+√3). Multiplication can be transformed into difference of two squares using the algebraic identity: (a-b)(a+b) = a^2 - b^2.

Where, we have to put in our expression a = √5 and b = √3.

→ [(3√5+√3)(√5+√3)]/[(√5)^2 - (√3)^3]

In denominator the square of √5 is 5. And the square of √3 is 3.

→ [(3√5+√3)(√5+√3)]/(5 - 3)

Subtract 3 from 5 to get 2.

[(3√5+√3)(√5+√3)]/2

Apply the distributive property by multiplying each term of 3√5+√3 by each term of √5 + √5 in numerator.

→ [3(√5)^2 + 3√5√3 + √3√5 + (√3)^2] / 2

The square of √5 is 5.

→ [3 × 5 + 2√5√3 + √3√5 + (√3)^2] / 2

Multiply 3 and 5 to get 15.

→ [15 + 3√5√3 + √3√5 + (√3)^2] / 2

To multiply √5 and √3, multiply the numbers under the square root.

→ [15 + 3√15 + √3√5 + (√3)^2] / 2

To multiply √3 and √5, multiply the numbers under the square root.

→ [15 + 3√15 + √15 + (√3)^2] / 2

Combine 3√15 and √15 to get 4√15.

→ [15 + 4√15 + (√3)^2] / 2

The square of √3 is 3.

→ (15 + 4√15 + 3)/2

Add 15 and 3 to get 18.

→ (18 + 4√15)/2

Divide each term of 18+4√15 by 2 to get 9+2√15.

→ 9 + 2√15.

Hence, the denominator is rationalised.

Answer:-

9 + 2√15.

Used Formula:-

Rationalising factor of √a + √b = √a - √b

(a-b)(a+b) = a^2 - b^2.