Question

$\tt \: If \: \: \: x + \frac{1}{x} = 20 \: \: find \: \: {x}^{2} + \frac{1}{ {x}^{2} }$

Answer 1

Answer:

Answer:

• 398

Given:

The equation is,

x+1/x=1

To find:

x²+1/x²=?

Formula used:

As we know that,

$( {a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab$

Solution:

The given equation is,

$x + \frac{1}{x} = 20$

Squaring on both sides,

x²+1/x²+2x×1/x=400

x²+1/x²+2=400

x²+1/x²=398

Answer 2

$\sf{\underline{\boxed{\red{\large{\bold{Question}}}}}}$

• $\sf \: If \: \: \: x + \frac{1}{x} = 20 \: \: find \: \: {x}^{2} + \frac{1}{ {x}^{2} }$

$\sf{\underline{\boxed{\red{\large{\bold{Solution}}}}}}$

• $\sf x + \frac{1}{x} = 20$

$\bf\underline{Squaring \ both \ sides:}$

$\sf{(x + \dfrac{1}{x})}^{2} = {(20)}^{2}$

Identify: $\sf{(a+b)^2=a^2+b^2+2ab}$

$\implies \: \sf {x}^{2} + { (\frac{1}{x} })^{2} + \: \: 2 \times x \times \frac{1}{x} = 400 \\ \\ \implies \sf {x}^{2} + \frac{1}{x^2} + 2 = 400 \\ \\ \implies \sf {x}^{2} + \frac{1}{x^2} = 400 - 2$

$\implies$${\underline {\boxed{\tt{x^2}{\dfrac{1}{x^2}=398}}}}$

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