Give a and b are rational numbers. Find a and b if:

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Give a and b are rational numbers. Find a and b if:

(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5

Solution:

(i) [3 – √5] / [3 + 2√5] = -19/11 + a√5
Let us consider LHS
[3 – √5] / [3 + 2√5]
Rationalize the denominator,
[3 – √5] / [3 + 2√5] = [(3 – √5) (3 – 2√5)] / [(3 + 2√5) (3 – 2√5)]
= [3(3 – 2√5) – √5(3 – 2√5)] / [32 – (2√5)2]
= [9 – 6√5 – 3√5 + 2.5] / [9 – 4.5]
= [9 – 6√5 – 3√5 + 10] / [9 – 20]
= [19 – 9√5] / -11
= -19/11 + 9√5/11
So when comparing with RHS
-19/11 + 9√5/11 = -19/11 + a√5
Hence, value of a = 9/11
(ii) [√2 + √3] / [3√2 – 2√3] = a – b√6
Let us consider LHS
[√2 + √3] / [3√2 – 2√3]
Rationalize the denominator,
[√2 + √3] / [3√2 – 2√3] = [(√2 + √3) (3√2 + 2√3)] / [(3√2 – 2√3) (3√2 + 2√3)]
= [√2(3√2 + 2√3) + √3(3√2 + 2√3)] / [(3√2)2 – (2√3)2]
= [3.2 + 2√2√3 + 3√2√3 + 2.3] / [9.2 – 4.3]
= [6 + 2√6 + 3√6 + 6] / [18 – 12]
= [12 + 5√6] / 6
= 12/6 + 5√6/6
= 2 + 5√6/6
= 2 – (-5√6/6)
So when comparing with RHS
2 – (-5√6/6) = a – b√6
Hence, value of a = 2 and b = -5/6
(iii) {[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]} = a + 7/11 b√5
Let us consider LHS
Since there are two terms, let us solve individually
{[7 + √5]/[7 – √5]}
Rationalize the denominator,
[7 + √5]/[7 – √5] = [(7 + √5) (7 + √5)] / [(7 – √5) (7 + √5)]
= [(7 + √5)2] / [72 – (√5)2]
= [72 + (√5)2 + 2.7.√5] / [49 – 5]
= [49 + 5 + 14√5] / [44]
= [54 + 14√5] / 44
Now,
{[7 – √5]/[7 + √5]}
Rationalize the denominator,
[7 – √5]/[7 + √5] = (7 – √5) (7 – √5)] / [(7 + √5) (7 – √5)]
= [(7 – √5)2] / [72 – (√5)2]
= [72 + (√5)2 – 2.7.√5] / [49 – 5]
= [49 + 5 – 14√5] / [44]
= [54 – 14√5] / 44
So, according to the question
{[7 + √5]/[7 – √5]} – {[7 – √5]/[7 + √5]}
By substituting the obtained values,
= {[54 + 14√5] / 44} – {[54 – 14√5] / 44}
= [54 + 14√5 – 54 + 14√5]/44
= 28√5/44
= 7√5/11
So when comparing with RHS
7√5/11 = a + 7/11 b√5
Hence, value of a = 0 and b = 1

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