Question

If a = root2+1 /root2-1 and b=root2-1/root2+1 then find the value of a 2+ b2

Answer 1

please find the attachment

Answer 2

Answer:

The value of ${a}^{2} + {b}^{2}$ is 34.

Step-by-step explanation:

Given,

$a = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1}$

and

$b = \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1 }$

So,

$a + b = \frac{ \sqrt{2} + 1 }{ \sqrt{2} - 1 } + \frac{ \sqrt{2} - 1 }{ \sqrt{2} + 1} \\ = \frac{ (\sqrt{2 } + 1) \times ( \sqrt{2 } + 1) + ( \sqrt{2 } - 1)( \sqrt{2} - 1) }{( \sqrt{2} - 1)( \sqrt{2} + 1) } \\ = \frac{ {( \sqrt{2} + 1) }^{2} + {( \sqrt{2} - 1) }^{2} }{ ({ \sqrt{2}) }^{2} - {1}^{2} } \\ = \frac{ { \sqrt{2} }^{2} + 2 \sqrt{2} + {1}^{2} + { \sqrt{2} }^{2} - 2 \sqrt{2} + 1 }{2 - 1} \\ = \frac{2 {( \sqrt{2} )}^{2} + 2 }{1} \\ = 2 \times 2 + 2 \\ = 4 + 2 \\ = 6$

and,

$a \times b = \frac{ \sqrt{2} + 1}{ \sqrt{2} - 1} \times \frac{ \sqrt{2} - + 1 }{ \sqrt{2} - 1} \\ = 1$

We know,

${a}^{2} + {b}^{2} \\ = {(a + b)}^{2} - 2ab \\ = {6}^{2} - 2 \times 1 \\ = 36 - 2 \\ = 34$

Here applied formulas,

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

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,

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

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

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