Question

if (x-1/x)=6, find the value of (x²+1/x²) and (x⁴+1/x⁴).​

Answer 1

Answer:

38 and 1442

Step-by-step explanation:

1.(x-1/x) = 6

squaring both side

x²+1/x² -2 = 36

x²+1/x² = 38

2. x²+1/x² = 38

squaring both side

x⁴+1/x⁴ +2 = 1444

x⁴+1/x⁴ = 1442

Answer 2

$\large\sf\underline{\underline{\red{Question}}}:$
$\:\:\: \red\maltese\: \sf If \: x - \frac{1}{x} = 6 , Find \:the\:value\:of\: \star x^2 + \frac{1}{x^2}\\ \:\:\:\:\:\: \star x^4 + \frac{1}{x^4}$

$\large\sf\underline{\underline{\red{Answer}}}:$
$\:\:\:\:\: \red\maltese \: \; x^2 + \frac{1}{x^2} = 38 \\\\ \:\:\:\:\: \red\maltese \:\: x^4 + \frac{1}{x^4} = 1442$
$\large\sf\underline{\underline{\red{Solution}}}:$

We have ,
$\mapsto \large x - \frac{1}{x} = 6$

$\textbf{Squaring both the sides}$

$\large \leadsto \left( x - \frac{1}{x} \right) ^2 = 6^2$

Using identity ,

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

$\implies x^2 + \left( \frac{1}{x} \right) ^2 - 2 \times \cancel{x} \times \left( \frac{1}{\cancel{x}} \right)= 6^2\\ \\ \implies x^2 + \frac{1}{x^2} - 2 = 36\\\\ \implies x^2 + \frac{1}{x^2} = 38 \\ \\ \large\implies \boxed{\red{x^2 + \frac{1}{x^2}= 38 }}$

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we have,

$\large \implies x^2 + \frac{1}{x^2} = 38$

$\textbf{Squaring both the sides}$
Using identity ,

$\red\bigstar\color{red} \boxed{\sf (a+b)^2 = a^2+b^2+2ab}$
$\implies (x^2)^2 + \left( \frac{1}{x^2 } \right) ^2 + 2 \times \cancel{x^2} \times \left( \frac{1}{\cancel{x^2}} \right)= 38^2\\ \\ \implies x^4 + \frac{1}{x^4} + 2 = 1444 \\\\ \implies x^4 + \frac{1}{x^4} = 1442 \\ \\ \large\implies \boxed{ \red{x^4 + \frac{1}{x^4}= 1442 }}$

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