Question

If a= 2-√3, and b=1/a, then find the value of a²+b²

Answer 1

Answer:

hope u understand the answer

Answer 2

Answer:

Required value of a²+b² is 14.

Step-by-step explanation:

Given,

$a = 2 - \sqrt{3}$and $b = \frac{1}{a} .....(1)$

Here we want to find value of

${a}^{2} + {b}^{2}$

We are putting value of a in equation (1),

$b = \frac{1}{2 - \sqrt{3} } \\ = \frac{2 + \sqrt{3} }{(2 - \sqrt{3})(2 + \sqrt{3}) } \\ = \frac{2 + \sqrt{3} }{ {2}^{2} - { \sqrt{3} }^{2} } \\ = \frac{2 + \sqrt{3} }{4 - 3} \\ = \frac{2 + \sqrt{3} }{1} \\ = 2 + \sqrt{3}$

Now,

${a}^{2} + {b}^{2} \\ = {(2 - \sqrt{3}) }^{2} + {(2 + \sqrt{3}) }^{2} \\ = {2}^{2} - 2.2. \sqrt{3} + { \sqrt{3} }^{2} + {2}^{2} + 2.2. \sqrt{3} + { \sqrt{3} }^{2} \\ = 4 - 4 \sqrt{3} + 3 + 4 + 4 \sqrt{3} + 3 \\ = 4 + 3 + 4 + 3 \\ = 14$

Here applied formula is ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$

and ${(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$

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,

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