Question

if x =7+root 48 find the value of x square + 1 upon x square​

Answer 1

Step-by-step explanation:

x= 7+√48

x^2 = (7)^2+(√48)^2 + 2× 7× √48

x^2= 49+48+14√48

x^2= 97+14√48

1/x^2= 97-14√48

so

X^2+1/X^2 = 97+14√48 + 97-14√48

= 97+97

= 194

Answer 2

The value of the expression $x^{2} +\frac{1}{x^{2} }$ is 194.

Given,

The value of x: $7+\sqrt{48}$.

To Find,

The value of the expression: $x^{2} +\frac{1}{x^{2} }$.

Solution,

The method of finding the value of the given expression is as follows -

First, we will compute the value of $x^{2}$.

$x^{2} =(7+\sqrt{48})^2=7^2+(\sqrt{48} )^2+2*7*\sqrt{48}$

$=49+48+14\sqrt{16*3}=97+14*4\sqrt{3}=97+56\sqrt{3}$.

Now we will compute the value of the expression $x^{2} +\frac{1}{x^{2} }$.

$x^{2} +\frac{1}{x^{2} }=97+56\sqrt{3}+ \frac{1}{97+56\sqrt{3}}$

$=97+56\sqrt{3}+\frac{97-56\sqrt{3}}{(97+56\sqrt{3})(97-56\sqrt{3})}$ [To rationalize the denominator]

$= 97+56\sqrt{3}+\frac{97-56\sqrt{3}}{(97)^2-(56\sqrt{3} )^2}$ [Since $(a+b)(a-b)=a^2-b^2$]

=$97+56\sqrt{3}+\frac{97-56\sqrt{3}}{9409-9408} =97+56\sqrt{3}+\frac{97-56\sqrt{3}}{1}$

$97+56\sqrt{3}+97-56\sqrt{3}=97+97=194$.

Hence, the value of the expression $x^{2} +\frac{1}{x^{2} }$ is 194.

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