Basics & LogarithmshardFree

Basics & Logarithms: Value Equal Prime Odd Integer Equal

JEE Maths question with a full step-by-step solution.

Question

x = \sqrt{4-2\sqrt{3}}

and

y = \sqrt{9-4\sqrt{5}}

, then the value of

(\sqrt{5}\,x - \sqrt{3}\,y)^2

is equal to

(a - b\sqrt{c})

, where

a, b, c

are co-prime and

c

is an odd integer. Then

a+b+c

is equal to:

36

correct

34

33

19

Solution

Step 1: Remove the radicals

4 - 2\sqrt{3} = 3 - 2\sqrt{3} + 1 = (\sqrt{3}-1)^2 \implies x = \sqrt{3}-1

9 - 4\sqrt{5} = 5 - 4\sqrt{5} + 4 = (\sqrt{5}-2)^2 \implies y = \sqrt{5}-2

(Both principal square roots are positive since

\sqrt{3} > 1

and

\sqrt{5} > 2

.) Step 2: Form

\sqrt{5}\,x - \sqrt{3}\,y

\sqrt{5}\,x = \sqrt{15}-\sqrt{5}, \quad \sqrt{3}\,y = \sqrt{15}-2\sqrt{3}

\sqrt{5}\,x - \sqrt{3}\,y = 2\sqrt{3} - \sqrt{5}

Step 3: Square the result

(2\sqrt{3}-\sqrt{5})^2 = 12 - 4\sqrt{15} + 5 = 17 - 4\sqrt{15}

Step 4: Identify the constants

a = 17

b = 4

c = 15

, which are pairwise co-prime with

c

odd.

a+b+c = 17+4+15 = 36

Answer: (1)

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