Question

x =(4-√5) find the value of (x² +1/x²)​

Answer 1

Answer:

Solution:-

• Given:

➝x=(4-√5(

• To find the value of:

➝(x²+1/x²)

• Here:

$➝ x = (4 - \sqrt{5} )$

On squaring both the sides:

$➝ {x}^{2} = {(4 - \sqrt{5)} }^{2}$

$➝ {x}^{2} = ( {4}^{2} ) + ( { \sqrt{5} }^{2} ) - 2 \times 4 \times \sqrt{5}$

$➝ {x}^{2} = 16 + 5 - 8 \sqrt{5}$

$➝ {x}^{2} = 21 + 8 \sqrt{5}$

Hence the value of, x²→21+8√5

• Again:

$➝ { (\frac{1}{x}) }^{2} = > \frac{1}{ {x}^{2} } = \frac{1}{21 + 8 \sqrt{5} }$

$➝ \frac{1}{ {x}^{2} } = \frac{1}{21 + 8 \sqrt{5} } \times \frac{21 - 8 \sqrt{5} }{21 - 5 \sqrt{5} }$

$➝ \frac{1}{ {x}^{2} } = \frac{21 - 5 \sqrt{5} }{( {21)}^{2} - ( {5 \sqrt{5}) }^{2} }$

$➝ \frac{1}{ {x}^{2} } = \frac{21 - 8 \sqrt{5} }{441 -( 64 \times 5)}$

$➝ \frac{1}{ {x}^{2} } = \frac{21 - 8 \sqrt{5} }{441 - 320}$

$➝ \frac{1}{ {x}^{2} } = \frac{21 - 8 \sqrt{5} }{121}$

• Therefore:

$➝ {x}^{2} + \frac{1}{ {x}^{2} } = 21 + 8 \sqrt{5} + \frac{21 - 5 \sqrt{5} }{121}$