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Basics & Logarithms: Natural Numbers Equal

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

Question

\sqrt{11+\sqrt{21}} = \sqrt{\dfrac{a}{b}} + \sqrt{\dfrac{c}{d}}

, where

a, b, c, d

are natural numbers with

\gcd(a,b) = \gcd(c,d) = 1

, then

a+b+c+d

is equal to:

\sqrt{21}

\sqrt{26}

26

correct

21

Solution

Step 1: Express the radical with a doubled surd

\sqrt{11+\sqrt{21}} = \sqrt{\frac{22+2\sqrt{21}}{2}} = \frac{\sqrt{22+2\sqrt{21}}}{\sqrt{2}}

Step 2: Denest the inner radical Seeking

m+n = 22

and

mn = 21

gives

m = 21

n = 1

22 + 2\sqrt{21} = (\sqrt{21}+1)^2 \implies \sqrt{22+2\sqrt{21}} = \sqrt{21}+1

Step 3: Write in the required form

\sqrt{11+\sqrt{21}} = \frac{\sqrt{21}+1}{\sqrt{2}} = \sqrt{\frac{21}{2}} + \sqrt{\frac{1}{2}}

Step 4: Identify the constants

a = 21

b = 2

c = 1

d = 2

, with

\gcd(21,2) = 1

and

\gcd(1,2) = 1

a+b+c+d = 21+2+1+2 = 26

Answer: (3)

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