CRE 4 - Comparison of Surds | Algebra - Surds & Indices

Answer: 1

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Explanation :

If a and b > 0 and a² > b², then a > b.

We have √6 + √3 and √7 + √2

Squaring both the terms

(6 + 3 + 2√18) and (7 + 2 + 2√14)

(9 + 2√18) and (9 + 2√14)

Since, rational part is same in both these terms, the term with greater irrational part will be greater, 

∴ 9 + 2√18 > 9 + 2√14

⇒ √6 + √3 > √7 + √2

Hence, 1.

Answer: 2

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Explanation :

If a and b > 0 and a² > b², then a > b.

We have √8 + √3 and √12 + √2

Squaring both the terms

(8 + 3 + 2√24) and (12 + 2 + 2√24)

(11 + 2√18) and (14 + 2√14)

Since, irrational part is same in both these terms, the term with greater rational part will be greater, 

∴ 14 + 2√24 > 11 + 2√24

⇒ √12 + √2 > √8 + √3

Hence, 2.

Answer: 1

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Explanation :

This is very straighforward question.

Here,

√5 > √4 and
√2 < √3

Hence, √5 - √2 > √4 - √3

In the first expression, a lower number √2 is bein subtracted from a higher number √5.

Hence, 1.

Answer: 1

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Explanation :

Assuming √6 - √3 < √10 - √7

⇒ √6 + √7 < √10 + √3

Squaring both sides, we get

(13 + 2√42) < (13 + 2√30)

But this is not true.

∴ Our assumption was wrong.

⇒ √6 - √3 > √10 - √7.

Hence, 1.

Answer: 2

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Explanation :

We have (√13 + √11) and (√15 + √10)

Squaring both sides we get

(24 + 2√143) and (25 + 2√150)

Since 25 > 24 and also 2√150 > 2√143

⇒ √15 + √10 > √13 + √11

Hence, 2.