Question

if
$x = 3 + \sqrt{8}$
then the value of
$( {x}^{2} + 1 \div {x}^{2} )$
is

​

Answer 1

Given

$\sf\to x = 3+\sqrt{8}$

To find

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

Now we can write as

$\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}$

Now rationalise

$\sf\to\dfrac{1}{x}=\dfrac{1}{3+\sqrt{8}}\times\dfrac{3-\sqrt{8}}{3-\sqrt{8}}=\dfrac{3-\sqrt{8}}{9-8} =3-\sqrt{8}$

we get

$\sf\to\dfrac{1}{x} =3-\sqrt{8}$

Using this identities

$\to\sf (a+b)^2= a^2+b^2+2ab$

Now we can write as

$\sf\to\bigg(x+\dfrac{1}{x}\bigg)^2=x^2+\dfrac{1}{x^2}+2\times x\times\dfrac{1}{x}$

Now put the value of x and 1/x

$\sf\to (3+\sqrt{8}+3-\sqrt{8})^2=x^2+\dfrac{1}{x^2} +2$

$\sf\to (3+3)^2=x^2+\dfrac{1}{x^2} +2$

$\sf\to (6)^2=x^2+\dfrac{1}{x^2} +2$

$\sf\to 36=x^2+\dfrac{1}{x^2} +2$

$\sf\to x^2+\dfrac{1}{x^2} =36-2$

$\sf\to x^2+\dfrac{1}{x^2} =34$

Answer

$\sf\to x^2+\dfrac{1}{x^2} =34$

Answer 2

${\bf{Given\::}}$

• $\bf{x = 3\: +\: \sqrt{8}}$

$\\ {\bf{To\: Find\::}}$

• $\rm{\left( {x}^{2} \:+ \:\dfrac{1}{ {x}^{2}} \right)}$

$\\ {\bf{Solution\::}}$

First we calculate the value of 'x²'.

➠ $\tt{\:x^2\:=\:\left(3\: +\: \sqrt{8}\right)^2}$

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

➠ $\bf{\:x^2\:=\:17\:+\:6\sqrt{8}\:}$

Now,

We calculate the value of $\rm{\left(\dfrac{1}{ {x}^{2}} \right)}.$

➠ $\tt{\:\dfrac{1}{ {(3\:+\:\sqrt{8})}^{2}} }$

➠ $\tt{\:\dfrac{1}{9\:+\:8\:+\:6\sqrt{8}} }$

➠ $\tt{\:\dfrac{1}{17\:+\:6\sqrt{8}} }$

Rationalize the above equation.

➠ $\tt{\:\dfrac{1}{17\:+\:6\sqrt{8}}\times \dfrac{17\:-\:6\sqrt{8}}{17\:-\:6\sqrt{8}} }$

➠ $\tt{\:\dfrac{17\:-\:6\sqrt{8}}{(17\:+\:6\sqrt{8})\:(17\:-\:6\sqrt{8})} }$

➠ $\tt{\:\dfrac{17\:-\:6\sqrt{8}}{(17)^2\:-\:(6\sqrt{8})^2} }$

➠ $\tt{\:\dfrac{17\:-\:6\sqrt{8}}{289\:-\:288} }$

➠ $\tt{\:\dfrac{17\:-\:6\sqrt{8}}{1} }$

➠ $\bf{\:(17\:-\:6\sqrt{8})}$

Now,

Adding these two results that we calculate.

$:\implies\:\rm{x^2\:+\:\dfrac{1}{ {x}^{2}} }$

$:\implies\:\rm{(17\:+\:6\sqrt{8})\:+\:(17\:-\:6\sqrt{8})\:}$

$:\implies\:\rm{17\:+\:\cancel{6\sqrt{8}}\:+\:17\:-\:\cancel{6\sqrt{8}}\:}$

$:\implies\:\rm{17\:+\:17\:}$

$:\implies\:\bf{34\:}$

⠀⠀⠀${\boxed{\bf{\pink{\therefore\:\left( {x}^{2} \:+ \:\dfrac{1}{ {x}^{2}} \right)\:=\:34\:}}}}$

__________________________

⠀⠀⠀⠀: SOME PROPERTIES :

⠀⠀❶ $\bf{(a\:+\:b)^2\:=\:a^2\:+\:b^2\:+\:2ab}$

⠀⠀❷ $\bf{(a\:+\:b)\:(a\:-\:b)\:=\:a^2\:-\:b^2\:}$