Question

if x = 3 + √8. find the value of x^2 + 1/x^2

Answer 1

x = 3+ √8

x = (3+√8)(3-√8) / (3-√8)

x = (9 -8) / (3-√8)

x = 1/ (3 -√8)

Or we can write 1/x = (3-√8)

Put the value and find out

x² +(1/x²) = x² + (1/x)²

So x² + 1/ x² = (3 +√8)² + (3 -√8)²

= 9+6√8+8 + 9–6√8+8

= 34. Ans.

Answer 2

$\bf {x}^{2} + \frac{1}{ {x}^{2} } = 34$

Given:

• $x = 3 + \sqrt{8}$

To find:

• Find the value of ${x}^{2} + \frac{1}{ {x}^{2} }$.

Solution:

Formula/Concept to be used:

• $\bf ( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$
• $\bf (a + b)(a - b) = {a}^{2} - {b}^{2}$

• Rationalization: It is used to free the denominator from any radical sign.
• For that, multiply and divide the fraction with RF of denominators.
• if denominator is $\bf a + \sqrt{b}$ then, RF factor is $\bf a - \sqrt{b}$

Step 1:

Squaring the x.

${x}^{2} = (3 + \sqrt{8})^{2}$

or

${x}^{2} = 9 + 8 + 6 \sqrt{8}$

or

${x}^{2} = 17 + 6 \sqrt{4\times 2}$

or

${x}^{2} = 17 + 6\times2 \sqrt{ 2}$

or

$\bf {x}^{2} = 17 + 12 \sqrt{2}$

Step 2:

Put the values in the expression.

${x}^{2} + \frac{1}{ {x}^{2} }$

or

$17 + 12 \sqrt{2} + \frac{1}{17 + 12 \sqrt{2} }$

rationalize the denominator.

$(17 + 12 \sqrt{2} ) + \frac{1}{17 + 12 \sqrt{2} } \times \frac{17 - 12 \sqrt{2} }{17 - 12 \sqrt{2} }$

or

$(17 + 12 \sqrt{2} ) + \frac{17 - 12 \sqrt{2} }{ {(17)}^{2} - ( {12 \sqrt{2} )}^{2} }$

or

$(17 + 12 \sqrt{2} ) + \frac{17 - 12 \sqrt{2} }{ 289 - 288}$

or

$(17 + 12 \sqrt{2} ) + \frac{17 - 12 \sqrt{2} }{1}$

or

$= 17 + 12 \sqrt{2} + 17 - 12 \sqrt{2}$

or

${x}^{2} + \frac{1}{ {x}^{2} } = 34$

Thus,

$\bf {x}^{2} + \frac{1}{ {x}^{2} } = 34$

Learn more:

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