The square root of {(8-4{3})}

Correct Answer is : (6−2)({6}-{2})(6−2)

Correct Answer is : (6−2)({6}-{2})(6−2)

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CGPSC Prelims CSAT 2025 Solved Paper 2

Question : 3 of 100

Marks: +1, -0

\sqrt{(8-4\sqrt{3})}

Solution:

Concept:

A binomial surd can be square-rooted by writing it as a perfect square of the form

(\sqrt{a}-\sqrt{b})^2

Explanation:

Let

8-4\sqrt{3}=(\sqrt{a}-\sqrt{b})^2

Expand:

(\sqrt{a}-\sqrt{b})^2=a+b-2\sqrt{ab}

Compare rational and irrational parts:

a+b=8

and

2\sqrt{ab}=4\sqrt{3}

\sqrt{ab}=2\sqrt{3}

, hence

ab=12

Solving

a+b=8

and

ab=12

gives

a=6

b=2

Therefore,

8-4\sqrt{3}=(\sqrt{6}-\sqrt{2})^2

Taking the principal square root:

\sqrt{8-4\sqrt{3}}=\sqrt{6}-\sqrt{2}

Answer:

Option C:

\sqrt{6}-\sqrt{2}

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