Question

Question:-
If, $\frac{30 - 5 \sqrt{3} }{ \sqrt{8} + \sqrt{5} + \sqrt{3} } = a + b \sqrt{3} + c \sqrt{5} + d \sqrt{10} + e \sqrt{15}$ (where a, b, c, d and e are rational numbers) then the value of (a + b + c + d + e) is:-
(A) 4
(B) 3
(C) 2
(D) 1

Answer with proper elucidation required!

Answer 1

Answer :

a + b + c + d + e + f = 4

⠀ ⠀⠀ ⠀ ⠀⠀ ⠀

Given :

$\bf \dfrac{30 - 5 \sqrt{3} }{ \sqrt{8} + \sqrt{5} + \sqrt{3} } = a + b \sqrt{3} + c \sqrt{5} + d \sqrt{10} + e \sqrt{15} + f \sqrt{30}$

⠀⠀ ⠀ ⠀⠀ ⠀

To Find :

• a + b + c + d + e + f = ?

⠀⠀ ⠀ ⠀⠀ ⠀

Solution :

⠀⠀ ⠀ ⠀⠀ ⠀

In LHS, we are given :

$\bf \dfrac{30 - 5 \sqrt{3} }{ \sqrt{8} + \sqrt{5} + \sqrt{3}}$

⠀⠀ ⠀ ⠀⠀ ⠀

By Rationalisation :

$\bf : \implies \dfrac{30 - 5 \sqrt{3} }{ \sqrt{8} + \sqrt{5} + \sqrt{3}} \times \dfrac{ \sqrt{8} - (\sqrt{5} + \sqrt{3})}{ \sqrt{8} - ( \sqrt{5} + \sqrt{3})}$

$\bf : \implies \dfrac{30\sqrt{8} - 30(\sqrt{5} + \sqrt{3} - 5 \sqrt{3} \times \sqrt{8} + 5 \sqrt{3} (\sqrt{5} + \sqrt{3})}{ (\sqrt{8})^{2} - (\sqrt{5} + \sqrt{3})^{2}}$

$\bf : \implies \dfrac{30 \times 2 \sqrt{2} - 30 \sqrt{5} - 30 \sqrt{3} - 5 \sqrt{3} \times 2 \sqrt{2} + 5 \sqrt{15} + 5 \sqrt{9}}{ (\sqrt{8})^{2} - (\sqrt{5} + \sqrt{3})^{2}}$

$\bf : \implies \dfrac{60 \sqrt{2} - 30 \sqrt{5} - 30 \sqrt{3} - 10 \sqrt{6} + 5 \sqrt{15} + 5 \times 3}{ 8 - ( 5 + 3 + 2 \sqrt{15})}$

$\bf : \implies \dfrac{60 \sqrt{2} - 30 \sqrt{5} - 30 \sqrt{3} - 10 \sqrt{6} + 5 \sqrt{15} + 15}{ 8 - 8 - 2 \sqrt{15}}$

$\bf : \implies \dfrac{60 \sqrt{2} - 30 \sqrt{5} - 30 \sqrt{3} - 10 \sqrt{6} + 5 \sqrt{15} + 15}{- 2 \sqrt{15}}$

⠀⠀ ⠀ ⠀⠀ ⠀

Now, again by rationalisation :

$\bf : \implies \dfrac{60 \sqrt{2} - 30 \sqrt{5} - 30 \sqrt{3} - 10 \sqrt{6} + 5 \sqrt{15} + 15}{- 2 \sqrt{15}} \times \dfrac{-2 \sqrt{15}}{-2 \sqrt{15}}$

$\bf : \implies \dfrac{- 120 \sqrt{30} + 60 \sqrt{5 \times 15} + 60 \sqrt{3 \times 15} + 20 \sqrt{6 \times 15} - 10 \times 15 - 30 \sqrt{15}}{(- 2 \sqrt{15})^{2}}$

$\bf : \implies \dfrac{- 120 \sqrt{30} + 60 \times 5 \sqrt{3} + 60 \times 3 \sqrt{5} + 20 \times 3 \sqrt{10} - 150 - 30 \sqrt{15}}{4 \times 15}$

$\bf : \implies \dfrac{- 120 \sqrt{30} + 300 \sqrt{3} + 180 \sqrt{5} + 60 \sqrt{10} - 150 - 30 \sqrt{15}}{60}$

$\bf : \implies \dfrac{30( - 4 \sqrt{30} + 10 \sqrt{3} + 6 \sqrt{5} + 2 \sqrt{10} - 5 - \sqrt{15})}{60}$

$\bf : \implies \dfrac{\cancel{30}( - 4 \sqrt{30} + 10 \sqrt{3} + 6 \sqrt{5} + 2 \sqrt{10} - 5 - \sqrt{15})}{\cancel{60}_{2}}$

$\bf : \implies \dfrac{- 4 \sqrt{30} + 10 \sqrt{3} + 6 \sqrt{5} + 2 \sqrt{10} - 5 - \sqrt{15}}{2}$

$\bf : \implies \dfrac{ -5 + 10 \sqrt{3} + 6 \sqrt{5} + 2 \sqrt{10} - \sqrt{15} - 4 \sqrt{30}}{2}$

$\bf : \implies \dfrac{ -5}{2} + \dfrac{ 10 \sqrt{3}}{2} + \dfrac{6 \sqrt{5}}{2} + \dfrac{2 \sqrt{10}}{2} - \dfrac{\sqrt{15}}{2} - \dfrac{4 \sqrt{30}}{2}$

$\bf : \implies \dfrac{ -5}{2} + \dfrac{ \cancel{10}^{5} \sqrt{3}}{\cancel{2}} + \dfrac{\cancel{6}^{3} \sqrt{5}}{\cancel{2}} + \dfrac{\cancel{2} \sqrt{10}}{\cancel{2}} - \dfrac{\sqrt{15}}{2} - \dfrac{\cancel{4}^{2} \sqrt{30}}{\cancel{2}}$

$\bf : \implies \dfrac{ -5}{2} + 5 \sqrt{3} + 3 \sqrt{5} + 1 \sqrt{10} - \dfrac{\sqrt{15}}{2} - 2 \sqrt{30}$

$\bf : \implies \dfrac{ -5}{2} + 5 \sqrt{3} + 3 \sqrt{5} + 1 \sqrt{10} + \Bigg( \dfrac{- 1}{2} \sqrt{15} \Bigg) + (- 2 \sqrt{30})$

⠀⠀ ⠀ ⠀⠀ ⠀

In RHS, we are given :

$\bf a + b \sqrt{3} + c \sqrt{5} + d \sqrt{10} + e \sqrt{15} + f \sqrt{30}$

⠀⠀ ⠀ ⠀⠀ ⠀

Now we are given LHS = RHS,

Hence, by comparing LHS with RHS, we get :

• $\bf a = \dfrac{ -5}{2}$

• $\bf b = 5$

• $\bf c = 3$

• $\bf d = 1$

• $\bf e = \dfrac{- 1}{2}$

• $\bf f = - 2$

⠀⠀ ⠀ ⠀⠀ ⠀

Now, we have to find a + b + c + d + e + f

$\bf = \dfrac{ -5}{2} + 5 + 3 + 1 + \Bigg(\dfrac{- 1}{2} \Bigg) + ( - 2)$

$\bf = \dfrac{ -5 - 1}{2} + 5 + 3 + 1 - 2$

$\bf = \dfrac{ - \cancel{6}^{3}}{\cancel{2}} + 7$

$\bf = - 3 + 7$

$\bf = 4$

⠀⠀ ⠀ ⠀⠀ ⠀

Hence, the answer is 4.