Question

. if x + 1/x =√5 ,Find the value of :
1. x^2 +1/x^2

if x^2 + 1/x^2 =83, find the value of x -1/x

Answer 1

$Given \: x + \frac{1}{x} = \sqrt{5} \: --(1)$

\* On squaring both sides of the equation, we get *\

$\Big( x + \frac{1}{x}\Big)^{2} =( \sqrt{5})^{2}$

$\implies x^{2} + \frac{1}{x^{2}} + 2 \times x \times \frac{1}{x} = 5$

$\implies x^{2} + \frac{1}{x^{2}} + 2 = 5$

$\implies x^{2} + \frac{1}{x^{2}} = 5 - 2$

$\green { \implies x^{2} + \frac{1}{x^{2}} = 3 }$

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Answer 2

Answer :

$\sf{x^2+\dfrac{1}{x^2}=3}$

Step-by-step explanation :

$\sf{x+\dfrac{1}{x}=\sqrt{5}}$

• Squaring on Both Sides

$\to\sf{\left(x+\dfrac{1}{x}\right)^2=(\sqrt{5})^2}$

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

$\to\sf{x^2+\dfrac{1^2}{x^2}+2\times x\times \dfrac{1}{x}=(\sqrt{5})^2}$

• We know $\sf{(\sqrt{a})^2=a}$

$\to\sf{x^2+\dfrac{1}{x^2}+2=5}$

• Subtracting 2 from both Sides

$\to\sf{x^2+\dfrac{1}{x^2}+2-2=5-2}$

$\boxed{\to\sf{x^2+\dfrac{1}{x^2}=3}}$