Question

please solve this question and please solve it with solution.​

Answer 1

${\huge{\bold{\pink{\bigstar{\boxed{\boxed{Question}}}}}}}$

If $x^2 + 4x - 1 = 0$ and x is positive than find the value of

• $x + \dfrac{1}{x}$
• $x^2 + \dfrac{1}{x^2}$

$\sf{\huge{\bold{\pink{\bigstar{\boxed{\boxed{Solution}}}}}}}$

$x^2 + 4x - 1 = 0$

• Divide the whole equation by x

$\dfrac{x^2}{x} + \dfrac{4x}{x} - \dfrac{1}{x} = \dfrac{0}{x}$

$x + 4 - \dfrac{1}{x} = 0$

$x + \dfrac{1}{x} = - 4$

$\sf{\bold{\red{\boxed{x + \dfrac{1}{x} = -4 }}}}$

$\sf{\bold{\green{\boxed{(x + \dfrac{1}{x})^2 = x^2 + \dfrac{1}{x^2} + 2 }}}}$

$\sf - 4 = x^2 + \dfrac{1}{x^2} + 2$

$\sf - 4 - 2 = x^2 + \dfrac{1}{x^2}$

$\sf x^2 + \dfrac{1}{x^2} = -6$

$\sf{\bold{\red{\boxed{x^2 + \dfrac{1}{x^2} = -6}}}}$

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