Question

3-√5/3+2√5 = a √(5- b/11)...... Find the values of a and b.

Answer 1

Answer:

hope it will help you thank you

Answer 2

Correct question is

$\frac{3 - \sqrt{5} }{3 + 2 \sqrt{5} } = a \sqrt{5} - \frac{b}{11}$

,find the values of a and b.

Answer:

Required value of a is $\frac{9}{19}$and value b is 11.

Step-by-step explanation:

Here

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

We are multiplying denominator and numerator by (3-2√5) .

$\frac{(3 - \sqrt{5})(3 - 2 \sqrt{5} ) }{(3 + 2 \sqrt{5} )(3 - 2 \sqrt{5} )} \\ = \frac{9 - 6 \sqrt{5} - 3 \sqrt{5} + 2 \times 5 }{ {3}^{2} - {(2 \sqrt{5} )}^{2} } \\ = \frac{9 - 9 \sqrt{5} + 10 }{9 - 20} \\ = \frac{19 - 9 \sqrt{5} }{ - 19} \\ = \frac{19}{ - 19} - \frac{9 \sqrt{5} }{ - 19} \\ = - 1 + \frac{9 \sqrt{5} }{19}$

According to question,

$- 1 + \frac{9 \sqrt{5} }{19} = a \sqrt{5} - \frac{b}{11}$

We are comparing both side,

$a = \frac{9}{19}$

$- 1 = - \frac{b}{11} \\ \frac{b}{11} = 1 \\ b = 11$

This is a problem of Algebra.

Some important Algebra formulas.

(a + b)² = a² + 2ab + b²

(a − b)² = a² − 2ab − b²

(a + b)³ = a³ + 3a²b + 3ab² + b³

(a - b)³ = a³ - 3a²b + 3ab² - b³

a³ + b³ = (a + b)³ − 3ab(a + b)

a³ - b³ = (a -b)³ + 3ab(a - b)

a² − b² = (a + b)(a − b)

a² + b² = (a + b)² − 2ab

a² + b² = (a − b)² + 2ab

a³ − b³ = (a − b)(a² + ab + b²)

a³ + b³ = (a + b)(a² − ab + b²)

Know more about Algebra,

1) https://brainly.in/question/13024124

2) https://brainly.in/question/1169549