Question

if X + 1 by X equal to 12 find the value of x minus one by X and X square + 1 by x square​

Answer 1

Answer:

please see the attachment

Answer 2

Given: $x+ \dfrac{1}{x} =12$

To find: $x-\dfrac{1}{x} \text { and } x^2+\dfrac{1}{x^2}$

Step-by-step explanation:

Now  as we know the algebraic identities

$(a+b)^2=a^2+b^2+2ab \text { and } (a-b)^2=a^2+b^2-2ab$

$x+ \dfrac{1}{x} =12\\\\ \text { Squaring both side }\\\\ (x+ \dfrac{1}{x})^2=12^2\\\\\Rightarrow x^2+\dfrac{1}{x^2}+2 \times x \times \dfrac{1}{x}=144\\\\\Rightarrow x^2+\dfrac{1}{x^2} +2=144\\\\\Rightarrow x^2+\dfrac{1}{x^2} =142$

Now

$(x-\dfrac{1}{x} )^2= x^2+\dfrac{1}{x^2} -2\times x \times \dfrac{1}{x}\\\\\Rightarrow (x-\dfrac{1}{x} )^2=142 -2\\\\\Rightarrow (x-\dfrac{1}{x} )^2=140 \\\\\Rightarrow x-\dfrac{1}{x} =\sqrt{140} = 2\sqrt{35}$

Hence, the value of $x-\dfrac{1}{x} =2\sqrt{35} \text { and } x^2+\dfrac{1}{x^2}=142$