Question

find the value of a and b in 3-√5/3+2√5=a√5-b where,a and b are rational number​

Answer 1

Answer:

What is the value of a and b, if (3+√5) / (3-√5) =a+b√5 where a and b are rational numbers?

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(3 + sqrt(5)) / (3 - sqrt(5)) = (a + b*sqrt(5)) /1 , we have one equation of 2 fractions.

The cross product gives: (3 + sqrt(5)) = (3 - sqrt(5)) * (a + b*sqrt(5))

Getting rid of the brackets: 3 + sqrt(5) = 3a + 3b*sqrt(5) - 5b - a* sqrt(5)

3 + sqrt(5) = (3a -5b) + (3b - a)*sqrt(5) Separating the Rational from the Irrational,

we get 2 linear equations: 3 = (3a -5b) and

1 = (3b - a)

Multiplying the second equation by 3 gives: 3 = - 3a + 9b

adding the first equation: 3 = 3a -5b

we get : 3+3= - 3a + 3a + 9b - 5b

Eliminating the variable a : 6 = 4 b this gives b = 6/4 = 3/2

Now knowing b=3/2 , we substitute it in 1 = (3b - a) and we get a = 7/2

The answer is: a = 7/2 ; and b = 3/2

Step-by-step explanation:

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Answer 2

Question:-

Find the value of a and b in

$\rm\frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a \sqrt{5} - b$

where a and b are rational number

Solution:-

$\rm \: \frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} }$

Now rationalize , we get

$\rm \: \frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } \times \frac{3 - 2 \sqrt{5} }{3 - 2 \sqrt{5} }$

Using this identity , we get

=> ( a - b )× ( c - d ) = ac - ad - bc + bd

=> ( a + b ) ( a - b ) = ( a² - b² )

$\rm \: \frac{(3 - \sqrt{5} )(3 - 2 \sqrt{5} )}{(3) {}^{2} - (2 \sqrt{5} ) {}^{2} }$

$\rm \: \frac{9 - 6 \sqrt{5 } - 3 \sqrt{5} + 2 \times 5 }{9 - 4 \times 5}$

$\rm \: \frac{19 - 9 \sqrt{5} }{ - 11}$

$\rm \: \frac{ - 19 + 9 \sqrt{5} }{11}$

$\rm \: \frac{9 \sqrt{5} - 19}{11}$

Now we can write as

$\rm \: \frac{9 \sqrt{5} }{11} - \frac{19}{11}$

$\rm \: value \: of \: a \: \: and \: \: b \: is$

$\rm \: a \: = \frac{9}{11} \: \: and \: \: b = \frac{19}{11}$