#### State which of the following are irrational numbers.

(i) 3 – √(7/25)

(ii) -2/3 + ∛2

(iii) 3/√3

(iv) -2/7 ∛5

(v) (2 – √3) (2 + √3)

(vi) (3 + √5)^{2}

(vii) (2/5 √7)^{2}

(viii) (3 – √6)^{2}

**Solution:**

(i) 3 – √(7/25)

Let us simplify,

3 – √(7/25) = 3 – √7/√25

= 3 – √7/5

Hence, 3 – √7/5 is an irrational number.

(ii) -2/3 + ∛2

Let us simplify,

-2/3 + ∛2 = -2/3 + 2^{1/3}

Since, 2 is not a perfect cube.

Hence it is an irrational number.

(iii) 3/√3

Let us simplify,

By rationalizing, we get

3/√3 = 3√3** **/(√3×√3)

= 3√3/3

= √3

Hence, 3/√3 is an irrational number.

(iv) -2/7 ∛5

Let us simplify,

-2/7 ∛5 = -2/7 (5)^{1/3}

Since, 5 is not a perfect cube.

Hence it is an irrational number.

(v) (2 – √3) (2 + √3)

Let us simplify,

By using the formula,

(a + b) (a – b) = (a)^{2} (b)^{2}

(2 – √3) (2 + √3) = (2)^{2} – (√3)^{2}

= 4 – 3

= 1

Hence, it is a rational number.

(vi) (3 + √5)^{2}

Let us simplify,

By using (a + b)^{2} = a^{2} + b^{2} + 2ab

(3 + √5)^{2} = 3^{2} + (√5)^{2} + 2.3.√5

= 9 + 5 + 6√5

= 14 + 6√5

Hence, it is an irrational number.

(vii) (2/5 √7)^{2}

Let us simplify,

(2/5 √7)^{2} = (2/5 √7) × (2/5 √7)

= 4/ 25 × 7

= 28/25

Hence it is a rational number.

(viii) (3 – √6)^{2}

Let us simplify,

By using (a – b)^{2} = a^{2} + b^{2} – 2ab

(3 – √6)^{2} = 3^{2} + (√6)^{2} – 2.3.√6

= 9 + 6 – 6√6

= 15 – 6√6

Hence it is an irrational number.

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