Question

if x-1/x=8 then the value of xsq.+1/xsq.​

Answer 1

Answer:

66

Step-by-step explanation:

$x - \frac{1}{x} = 8$

Squaring both sides,

$(x - \frac{1}{x} )^{2} = 8^{2}$

$x^{2} + \frac{1}{x^{2} } -2(x)(\frac{1}{x} ) = 64$

$x^{2} + \frac{1}{x^{2} } -2 = 64$

$x^{2} + \frac{1}{x^{2} } = 66$

Hope this answers your question!! :)

Answer 2

Given :-

x-1/x= 8

To find :-

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

identity used

$\boxed{\sf{(a-b){}^{2}=a{}^{2}+b{}^{2}-2ab}}$

Now

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

put the value of x-1/x

$\sf\bullet \bigg(x-\dfrac{1}{x}\bigg){}^{2}=x{}^{2}+\dfrac{1}{x{}^{2}}-2\\ \\ \sf\implies (8){}^{2}=x{}^{2}+\dfrac{1}{x{}^{2}}-2\\ \\ \sf\implies 64+2=x{}^{2}+\dfrac{1}{x{}^{2}}\\ \\ \sf\implies 66=x{}^{2}+\dfrac{1}{x{}^{2}}$

$\boxed{\tt{\blue{x{}^{2}+\dfrac{1}{x{}^{2}}=66}}}$