Question

if a=1- √5 , find the value of a² - 1/a²​

Answer 1

Answer:

Value of ${a}^{2} - \frac{1}{ {a}^{2} }$

is $\frac{45 - 17 \sqrt{5} }{8}$

Step-by-step explanation:

Given,$a = 1 - \sqrt{5}$

Here we want to find value of ${a}^{2} - \frac{1}{ {a}^{2} }$

Now,

${a}^{2} = {(1 - \sqrt{5}) }^{2} \\ = {1}^{2} - 2.1. \sqrt{5} + { \sqrt{5} }^{2} \\ = 1 - 2 \sqrt{5} + 5 \\ = 6 - 2 \sqrt{5}$

and

$\frac{1}{ {a}^{2} } = \frac{1}{6 - 2 \sqrt{5} } \\ = \frac{(6 + 2 \sqrt{5} )}{(6 - 2 \sqrt{5} )((6 + 2 \sqrt{5} )} \\ = \frac{(6 + 2 \sqrt{5} )}{ {6}^{2} - {(2 \sqrt{5} )}^{2} } \\ = \frac{(6 + 2 \sqrt{5} )}{36 - 20} \\ = \frac{2(3 + \sqrt{5} )}{16} \\ = \frac{3 + \sqrt{5} }{8}$

Now,

${a}^{2} - \frac{1}{ {a}^{2} } \\ = 6 - 2 \sqrt{5} - \frac{3 + \sqrt{5} }{8} \\ = \frac{48 - 16 \sqrt{5} - 3 - \sqrt{5} }{8} \\ = \frac{45 - 17 \sqrt{5} }{8}$

Here applied formula,

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

This is a problem of Algebra.

Some important Algebra formulas are

(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

Answer 2

Answer:

These is the correct answer of these question.

Step-by-step explanation:

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