Question

if a=√5+2/√5-2,b=√5-2/√5+2 then a+b =​

Answer 1

Answer:

18

Step-by-step explanation:

As per the provided information in the given question, we have to calculate the value of (a + b) where,

⠀⠀⠀• $\sf { a = \dfrac{\sqrt{5}+ 2}{\sqrt{5}-2 } }$

⠀⠀⠀• $\sf { b = \dfrac{\sqrt{5}- 2}{\sqrt{5}+2 } }$

In order to calculate the value of (a + b), firstly we need to rationalize the denominator of a and b.

In order to rationalise the denominator, we multiply the rationalising factor of the denominator with both the numerator and the denominator of the fraction.

Rationalising the denominator of a :

$\twoheadrightarrow \sf{\quad {a = \dfrac{\sqrt{5}+ 2}{\sqrt{5}-2} }}$

Here,the denominator is in the form of (a – b) and the rationalising factor of (a – b) is (a + b). So, the rationalising factor of (√5 – 2) is (√5 + 2). Multiplying (√5 + 2) with both the numerator and the denominator of the fraction.

$\twoheadrightarrow \sf{\quad {a = \dfrac{\sqrt{5}+ 2}{\sqrt{5}-2}\times \dfrac{\sqrt{5}+ 2}{\sqrt{5}+2} }}$

As we know the fraction rules that, a/b × c/d = ac/ad. So, we can rearrange the terms as,

$\twoheadrightarrow \sf{\quad {a = \dfrac{(\sqrt{5}+ 2)^2}{(\sqrt{5}-2)(\sqrt{5}+2)} }}$

By using the identities :

⠀⠀⠀⠀⠀★ (a + b)² = a² + b² + 2ab

⠀⠀⠀⠀⠀★ (a + b)(a – b) = a² – b²

$\twoheadrightarrow \sf{\quad {a = \dfrac{(\sqrt{5})^2 + (2)^2 + 2(2\sqrt{5}) }{(\sqrt{5})^2-(2)^2} }}$

Writing the squares of the terms in the numerator and the denominator and performing multiplication in the numerator.

$\twoheadrightarrow \sf{\quad {a = \dfrac{5 + 4 + 4\sqrt{5} }{5 - 4} }}$

Performing addition in the numerator and performing subtraction in the denominator.

$\twoheadrightarrow \quad\underline{\boxed { \pmb{\frak{a = 9 + 4\sqrt{5}}} }}$

Now, we have to rationalise the denominator of the b, rationalising the denominator of b.

Rationalising the denominator of b :

$\twoheadrightarrow \sf{\quad {a = \dfrac{\sqrt{5}-2}{\sqrt{5}+2} }}$

Here,the denominator is in the form of (a + b) and the rationalising factor of (a + b) is (a – b). So, the rationalising factor of (√5 + 2) is (√5 – 2). Multiplying (√5 – 2) with both the numerator and the denominator of the fraction.

$\twoheadrightarrow \sf{\quad {b = \dfrac{\sqrt{5}- 2}{\sqrt{5}+2}\times \dfrac{\sqrt{5}- 2}{\sqrt{5}-2} }}$

As we know the fraction rules that, a/b × c/d = ac/ad. So, we can rearrange the terms as,

$\twoheadrightarrow \sf{\quad {b = \dfrac{(\sqrt{5}- 2)^2}{(\sqrt{5}+2)(\sqrt{5}-2)} }}$

By using the identities :

⠀⠀⠀⠀⠀★ (a – b)² = a² + b² – 2ab

⠀⠀⠀⠀⠀★ (a + b)(a – b) = a² – b²

$\twoheadrightarrow \sf{\quad {b = \dfrac{(\sqrt{5})^2 + (2)^2 - 2(2\sqrt{5}) }{(\sqrt{5})^2-(2)^2} }}$

Writing the squares of the terms in the numerator and the denominator and performing multiplication in the numerator.

$\twoheadrightarrow \sf{\quad {b= \dfrac{5 + 4 - 4\sqrt{5} }{5 - 4} }}$

Performing addition in the numerator and performing subtraction in the denominator.

$\twoheadrightarrow \quad\underline{\boxed { \pmb{\frak{b = 9 - 4\sqrt{5}}} }}$

⠀⠀⠀_________________________⠀⠀⠀⠀⠀

❝ Value of (a + b) : ❞

$\twoheadrightarrow \sf{\quad {(a + b) }}$

Substitute the value of a and b.

$\twoheadrightarrow \sf{\quad { (9 + 4\sqrt{5})+ (9 - 4\sqrt{5}) }}$

Removing the brackets.

$\twoheadrightarrow \sf{\quad { 9 + 4\sqrt{5}+ 9 - 4\sqrt{5} }}$

Performing addition and subtraction of the terms.

$\twoheadrightarrow \quad\underline{\boxed { \pmb{\frak{(a + b) = 18 }} }}$

❝ Therefore, the value of (a + b) is 18. ❞