Question

if a=8+3 root7,find value of a^2+1/a^2

Answer 1

Answer:

254

Step-by-step explanation:

a = 8 + 3√7

a² = ( 8 + 3√7 )² = 8² + (3√7)² + 2×8×3√7

   = 64 + 63 + 48√7

   = 127 + 48√7

1/a² = 1 / ( 127 + 48√7 )

      = ( 127 - 48√7 ) / ( ( 127 + 48√7 ) ( 127 - 48√7 ) )

      = ( 127 - 48√7 ) / ( 127² - (48√7)² )

      = ( 127 - 48√7 ) / ( 16129 - 16128 )

      = 127 - 48√7

a² + 1/a² = ( 127 + 48√7 ) + ( 127 - 48√7 ) = 127 + 127 = 254

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Or a better, slightly trickier, way:

a = 8 + 3√7

1/a = 1 / ( 8 + 3√7 )

    = ( 8 - 3√7 ) / ( ( 8 + 3√7 ) ( 8 - 3√7 ) )

    = ( 8 - 3√7 ) / ( 8² - (3√7)² )

    = ( 8 - 3√7 ) / ( 64 - 63 )

    = 8 - 3√7

a + 1/a = ( 8 + 3√7 ) + ( 8 - 3√7 ) = 8 + 8 = 16

a² + 1/a² = ( a + 1/a )² - 2 = 16² - 2 = 256 - 2 = 254

Hope that helps!

Answer 2

Answer:

254

Step-by-step explanation:

Given that,

a = 8 + 3√7

To find the value of a^2 + 1/a^2.

First of all, we need to find the value of a^2.

$= > {a}^{2} = {(8 + 3 \sqrt{7}) }^{2}$

But, we know that,

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

Therefore, we will get,

$= > {a}^{2} = {(8)}^{2} + 2(8)(3 \sqrt{7} ) + {(3 \sqrt{7} )}^{2} \\ \\ = > {a}^{2} = 64 + 48 \sqrt{7} + 63 \\ \\ = > {a}^{2} = 127 + 48 \sqrt{7}$

Therefore, we will get,

$= > \frac{1}{ {a}^{2} } = \frac{1}{127 + 48 \sqrt{7} }$

Now, we have,

$a = 8 + 3 \sqrt{7}$

Therefore, we have,

$= > \frac{1}{a} = \frac{1}{8 + 3 \sqrt{7} } \\ \\ = > \frac{1}{a} = \frac{1}{8 + 3 \sqrt{7} } \times \frac{8 - 3 \sqrt{7} }{8 - 3 \sqrt{7} } \\ \\ = > \frac{1}{a} = \frac{8 - 3 \sqrt{7} }{ {(8)}^{2} - {(3 \sqrt{7} )}^{2} } \\ \\ = > \frac{1}{a} = \frac{8 - 3 \sqrt{7} }{64 - 63} \\ \\ = > \frac{1}{a} = \frac{8 - 3 \sqrt{7} }{ 1} \\ \\ = > \frac{1}{a} = 8 - 3 \sqrt{7}$

Now, we have,

$= > {a}^{2} + \frac{1}{ {a}^{2} } = {(a + \frac{1}{a} )}^{2} - 2$

Substituting the values, we get,

$= > {a}^{2} + \frac{1}{ {a}^{2} } ={ (8 + 3 \sqrt{7} + 8 - 3 \sqrt{7} ) }^{2} - 2 \\ \\ = > {a}^{2} + \frac{1}{ {a}^{2} } = {(16)}^{2} - 2 \\ \\ = > {a}^{2} + \frac{1}{ {a}^{2} } = 256 - 2 \\ \\ = > {a}^{2} + \frac{1}{ {a}^{2} } = 254$

Hence, required value is 254.