Question

Equals to 3 + 2 root 2 then find the value of a square + 1 upon a square

Answer 1

Answer:

34.

Step-by-step explanation:

Given,

a = 3 + 2√2

$\frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} }$

Rationalizing the denominator we get:

$\frac{1}{a} = \frac{1}{3 + 2 \sqrt{2} } \times \frac{3 - 2 \sqrt{2} }{3 - 2 \sqrt{2} } \\ \\ \frac{1}{a} = \frac{3 - 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2} )}^{2} } \\ \\ \frac{1}{a} = \frac{3 - 2 \sqrt{2} }{9 - 8} \\ \\ \bf \frac{1}{a} = 3 - 2 \sqrt{2}$

Now,

$a + \frac{1}{a} = (3 + 2 \sqrt{2} ) + (3 - 2 \sqrt{2} ) \\ \\ a + \frac{1}{a} = 3 + 2 \sqrt{2} + 3 - 2 \sqrt{2} \\ \\ a + \frac{1}{a} = 6$

Squaring both the sides we get:

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

Answer 2

a = 3 + 2√2

______ [GIVEN]

• We have to find the value of a² + $\dfrac{1}{ {a}^{2} }$

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Solution:

• a = 3 + 2√2

• $\dfrac{1}{a}$ = $\dfrac{1}{3 \: + \: 2 \sqrt{2} }$

» Rationalize

=> $\dfrac{1}{3 \: + \: 2 \sqrt{2} }$ × $\dfrac{3 \: - \: 2 \sqrt{2} }{3 \: - \: 2\sqrt{2} }$

(a + b) (a - b) = a² - b²

=> $\dfrac{3 \: - \: 2 \sqrt{2} }{ {3}^{2} \: - \: 4 \: \times \: 2 }$

=> $\dfrac{3\:-\:2\sqrt{2}}{9\:-\:8}$

=> 3 - 2√2

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Now..

=> (a + $\dfrac{1}{a})$ = 3 + 2√2 + 3 - 2√2

=> (a + $\dfrac{1}{a})$ = 6

• Squaring on both sides.

=> $( {a \: + \: \dfrac{1}{a} )}^{2}$ = (6)²

(a + b)² = a² + b² + 2ab

=> a² + $\dfrac{1}{ {a}^{2} }$ + 2a × $\dfrac{1}{a}$ = 36

=> a² + $\dfrac{1}{ {a}^{2} }$ + 2 = 36

=> a² + $\dfrac{1}{ {a}^{2} }$ = 36 - 2

_____________________________

a² + $\dfrac{1}{ {a}^{2} }$ = 34

___________ [ANSWER]