Given that $y = \frac{x^2}{\sqrt{1 - x - x^2}},$ find the range of values of $x$ for \

Question

Given that $y = \frac{x^2}{\sqrt{1 - x - x^2}},$ find the range of values of $x$ for $y$ to be valid, and prove that $2y \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{x^3 (2x^2 + 3x - 4)}{(1 - x - x^2)^2} = 0.$

Accepted Answer

(i) $\dfrac{-1-\sqrt5}{2}<x<\dfrac{-1+\sqrt5}{2}$ (ii) $2y\dfrac{\mathrm{d}y}{\mathrm{d}x}+\dfrac{x^3(2x^2+3x-4)}{(1-x-x^2)^2}=0$

Quadratic Functions, Equations and Inequalities - Quadratic Inequalities

Quadratic Functions, Equations and Inequalities - Quadratic Inequalities

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