Question

if x-1/x=6, find the value of
a. x²+1/x²
b. x⁴+1/x⁴​

Answer 1

$\huge\mathbb{SOLUTION:-}$

• Given the value of
• $\sf x+\frac{1}{x}=6$

We have to find :-

$\sf\leadsto x{}^{2}+\frac{1}{x{}^{2}}\\ \\ \sf\leadsto x{}^{4}+\frac{1}{x{}^{4}}$

Now :-

$\bf\underline{\red{According\:to\: Question:-}}$

Formula used :-

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

• In the place of a = x

• and b= 1/x

$\sf (x+\frac{1}{x}){}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2\times \cancel{x}+\frac{1}{\cancel{x}}\\ \\ \implies{\blue{\mathfrak{(x+\frac{1}{x}){}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2}}}$

$\sf\implies (x+\frac{1}{x}){}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2\\ \\ \sf\implies (6){}^{2}=x{}^{2}+\frac{1}{x{}^{2}}+2\\ \\ \sf\implies 36-2=x{}^{2}+\frac{1}{x{}^{2}}\\ \\ \sf\implies 34=x{}^{2}+\frac{1}{x{}^{2}}$

$\boxed{\mathfrak{\star{\purple{x{}^{2}+\frac{1}{x{}^{2}}=34}}}}$

• Now the value of
• $\sf\leadsto x{}^{4}+\frac{1}{x{}^{4}}$

$\sf (x{}^{2}+\frac{1}{x{}^{2}}){}^{2}=x{}^{4}+\frac{1}{x{}^{4}}+2\times \cancel{x{}^{2}}+\frac{1}{\cancel{x{}^{2}}}\\ \\ \implies {\purple{\mathfrak{(x{}^{2}+\frac{1}{x{}^{2}}){}^{2}=x{}^{4}+\frac{1}{x{}^{4}}+2}}}$

$\sf\implies (x{}^{2}+\frac{1}{x{}^{2}}){}^{2}=x{}^{4}+\frac{1}{x{}^{4}}+2\\ \\ \sf\implies (34){}^{2}=x{}^{4}+\frac{1}{x{}^{4}}+2\\ \\ \sf\implies 1156-2=x{}^{4}+\frac{1}{x{}^{4}}\\ \\ \sf\implies 1154=x{}^{4}+\frac{1}{x{}^{4}}$

$\boxed{\mathfrak{\star{\purple{x{}^{2}+\frac{1}{x{}^{2}}=34}}}}$

$\boxed{\mathfrak{\star{\purple{x{}^{4}+\frac{1}{x{}^{4}}=1154}}}}$