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If a :b=√7 : √3, then the value of (3a+2b) :(3a-2b) is equal to:
Question

If a :b=√7 : √3, then the value of (3a+2b) :(3a-2b) is equal to:

A.

(2+21)(221)\frac{(2+√21)}{(2-√21) }​​

B.

(2+21)(2+21)\frac{(2+√21)}{(-2+√21) }​​

C.

(2+21)(221)\frac{(2+√21)}{(-2-√21)}​​

D.

(221)(2+21)\frac{(2-√21)}{(2+√21)}​​

Correct option is B

Given:
a:b=7:3a : b = \sqrt{7} : \sqrt{3} \\​​
Formula used:
(a+b)2=a2+b2+2ab(a2b2)=(a+b)(ab)(a + b)^2 = a^2 + b^2 + 2ab \\(a^2 - b^2) = (a + b)(a - b) \\​​
Solution:  
Let a=7x and b=3x3×7x+2×3x3×7x2×3x=(3×7x+2×3x)×(3×7x+2×3x)(3×7x2×3x)×(3×7x+2×3x)=(3×7x)2+(2×3x)2+2×(37x)(23x)(37x)2(23x)2=63x2+12x2+1221x263x212x2=75+122151=25+42117Check option (b)(2+21)(2+21)Rationalize by multiplying (2+21) in both numerator and denominator=(2+21)2(2+21)(2+21)=25+42117\text{Let } a = \sqrt{7x} \text{ and } b = \sqrt{3x}\newline\frac{3 \times \sqrt{7x} + 2 \times \sqrt{3x}}{3 \times \sqrt{7x} - 2 \times \sqrt{3x}}\newline= \frac{(3 \times \sqrt{7x} + 2 \times \sqrt{3x}) \times (3 \times \sqrt{7x} + 2 \times \sqrt{3x})}{(3 \times \sqrt{7x} - 2 \times \sqrt{3x}) \times (3 \times \sqrt{7x} + 2 \times \sqrt{3x})}\newline= \frac{(3 \times \sqrt{7x})^2 + (2 \times \sqrt{3x})^2 + 2 \times (3\sqrt{7x})(2\sqrt{3x})}{(3\sqrt{7x})^2 - (2\sqrt{3x})^2}\newline= \frac{63x^2 + 12x^2 + 12\sqrt{21}x^2}{63x^2 - 12x^2}\newline= \frac{75 + 12\sqrt{21}}{51}\newline= \frac{25 + 4\sqrt{21}}{17}\newline\text{Check option (b)}\newline\frac{(2 + \sqrt{21})}{(-2 + \sqrt{21})}\newline\text{Rationalize by multiplying } (2 + \sqrt{21}) \text{ in both numerator and denominator}\newline= \frac{(2 + \sqrt{21})^2}{(-2 + \sqrt{21})(2 + \sqrt{21})}\newline= \frac{25+4\sqrt{21}}{17}​​
Thus, correct option is (b).

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