Question

if x = 3-2√2 find x2 + 1/x2​

Answer 1

Answer:

34

Step-by-step explanation:

$x = 3 - 2 \sqrt{2}$

To find ,

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

Using identity (a+b}^2

(x+1/x)^2=x^2+1/x^2 +2

x^2+1/x^2 = (x+1/x)^2-2

${x}^{2} + \frac{1}{ {x}^{2} } = (3 - 2 \sqrt{2 } + \frac{1}{3 - 2 \sqrt{2} } )^{2} - 2$

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

${6}^{2} - 2$

$= 36 - 2 = 34$

Answer 2

Answer:

34

Step-by-step explanation:

$\large \text{Given x=3-2\sqrt{2} }\\\\\\\large \text{we have to find value of x^2+\dfrac{1}{x^2} }\\\\\\\large \text{Taking reciprocal of x both side }\\\\\\\large \text{ \dfrac{1}{x}= \dfrac{1}{3-2\sqrt{2} } }\\\\\\\large \text{Now rationalize the denominator}\\\\\\\large \text{ \dfrac{1}{x}= \dfrac{1}{3-2\sqrt{2}}\times \dfrac{3+2\sqrt{2}}{3+2\sqrt{2}} }\\\\\\\large \text{ \dfrac{1}{x}=\dfrac{3+2\sqrt{2}}{9-8} }\\\\\\\large \text{ \dfrac{1}{x}={3+2\sqrt{2}} }$

$\large \text{Now adding both}\\\\\\\large x+\dfrac{1}{x}=3-2\sqrt{2}+3+2\sqrt{2} \\\\\\\large x+\dfrac{1}{x}=6\\\\\\\large \text{Now squaring on both side}\\\\\\\large (x+\dfrac{1}{x})^2=(6)^2\\\\\\\large x^2+\dfrac{1}{x^2}+2=36\\\\\\\large x^2+\dfrac{1}{x^2}=36-2\\\\\\\large x^2+\dfrac{1}{x^2}=34\\\\\\\large \text{Thus we get answer 34}$