Question

If x = V2 +1, find the value of x² +1/ x^2​

Answer 1

6

Step-by-step explanation:

${\underline{{\text{Given}}}}$

$x = \sqrt{2} + 1$

${\underline{{\text{To Find:}}}}$

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

${\underline{{\text{Solution:}}}}$

1st we should find the value of $x^2 \: \text{and} \dfrac{1}{x^2}$

So,

$x = \sqrt{2} + 1 \\ \\ {x}^{2} = {( \sqrt{2} + 1)}^{2} \: \: \: \: [\text{Using...1}] \\ \\ = {( \sqrt{2}) }^{2} + {(1)}^{2} + 2. \sqrt{2} .1 \\ \\ = 2 + 1 + 2 \sqrt{2} \\ \\ \therefore x {}^{2} = 3 + 2 \sqrt{2}$

Again,

${x}^{2} = 3 + 2\sqrt{2} \\ \\ \frac{1}{x {}^{2} } = \frac{1}{3 + 2\sqrt{2} } \\ \\ = \frac{(3 - 2\sqrt{2} )}{( 3 + 2\sqrt{2})( 3 - 2\sqrt{2} ) } \\ \\ [\text{Rationalizing the denominator}] \\ \\ = \frac{3 - 2\sqrt{2} }{ { {(3)}^{2} - (2\sqrt{2} )}^{2} } \: \: \: [\text{Using....2 }] \\ \\ = \frac{3 - 2\sqrt{2} }{9 - 8} \\ \\ = \frac{3 - 2 \sqrt{2} }{1} \\ \\ \therefore \: \frac{1}{ {x}^{2} } = 3 - 2 \sqrt{2}$

Now,

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

Hence, the required answer is 6.

${\underline{{\text{ Using Algebra Formula }}}}$

${(x + y)}^{2}={x}^{2}+{y}^{2}+2xy.....(1)\\ \\(x+y)(x-y)=x^2-y^2......(2) \\ \\{(x - y)}^{2}={x}^{2}+{y}^{2}-2xy\\ \\{(x+y)}^{2}= (x - y) {}^{2}+4xy\\ \\{(x-y)}^{2}=(x+y){}^{2}-4xy\\ \\ (x + y)^{2}+(x-y)^{2}=2( {x}^{2}+{y}^{2} )\\ \\(x+y)^{2}- (x-y) {}^{2}=4xy\\ \\ {(x + y)}^{3}={x}^{3}+{y}^{3}+ 3xy(x + y) \\ \\(x - y)^{3}={x}^{3}-{y}^{3}- 3xy(x - y)$

Answer 2

$\bf Given:\ \tt x\ =\ \sqrt{2}\ +\ 1.\\\\\\\bf To\ find:\ \tt The\ value\ of\ x^2\ +\ \dfrac{1}{x^2}.\\\\\\\bf Answer:$

$\sf Since\ we\ know\ the\ value\ of\ x\ [\sqrt{2}\ +\ 1],\ let's\ find\ its\ square.\\\\\\\tt x^2\ \implies\ (\sqrt{2}\ +\ 1)^2\ =\ 2\ +\ 2\sqrt{2}\ +\ 1\ =\ \bf 3\ +\ 2\sqrt{2}.$

$\sf This\ means\ that\ \tt \dfrac{1}{x^2}\ \implies\ \dfrac{1}{3\ +\ \2\sqrt{2}}.\\\\\\\sf Rationalizing\ it,\\\\\\\implies\ \tt \dfrac{1}{3\ +\ 2\sqrt{2}}\\\\\\=\ \dfrac{1(3\ -\ 2\sqrt{2})}{(3\ +\ 2\sqrt{2})(3\ -\ 2\sqrt{2})}\\\\\\=\ \dfrac{3\ -\ 2\sqrt{2}}{(3)^2\ -\ (2\sqrt{2})^2}\\\\\\=\ \dfrac{3\ -\ 2\sqrt{2}}{9\ -\ 8}\\\\\\=\ \bf 3\ -\ 2\sqrt{2}$

Now that we have both the values, let's substitute them and solve it.

$\tt x^2\ +\ \dfrac{1}{x^2}\ \implies\ 3\ +\ 2\sqrt{2}\ +\ 3\ -\ 2\sqrt{2}\\\\\\\bf x^2\ +\ \dfrac{1}{x^2}\ =\ 6$