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(D) For equation $I$: $2x^{2} + x - 1 = 0$Using the quadratic formula or factoring: $2x^{2} + 2x - x - 1 = 0 \Rightarrow 2x(x + 1) - 1(x + 1) = 0 \Rig

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if $x \le y$

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(D) For equation $I$: $2x^{2} + x - 1 = 0$Using the quadratic formula or factoring: $2x^{2} + 2x - x - 1 = 0 \Rightarrow 2x(x + 1) - 1(x + 1) = 0 \Rightarrow (2x - 1)(x + 1) = 0$.Thus,the roots are $x = \frac{1}{2}$ and $x = -1$.For equation $II$: $2y^{2} - 3y + 1 = 0$Factoring: $2y^{2} - 2y - y + 1 = 0 \Rightarrow 2y(y - 1) - 1(y - 1) = 0 \Rightarrow (2y - 1)(y - 1) = 0$.Thus,the roots are $y = \frac{1}{2}$ and $y = 1$.Comparing the values:If $x = \frac{1}{2}$,then $y$ can be $\frac{1}{2}$ (equal) or $1$ $(x If $x = -1$,then $y$ can be $\frac{1}{2}$ or $1$ $(x In all cases,$x \le y$.

Solve the given two equations and select the correct answer from the given options.
$I.$ $2x^{2} + x - 1 = 0$
$II.$ $2y^{2} - 3y + 1 = 0$

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Let $p, q$ and $r$ be real numbers $(p \ne q, r \ne 0),$ such that the roots of the equation $\frac{1}{x + p} + \frac{1}{x + q} = \frac{1}{r}$ are equal in magnitude but opposite in sign. Then the sum of squares of these roots is equal to:

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Solve the given two equations and select the correct answer from the given options.$I.$ $2x^{2} + x - 1 = 0$$II.$ $2y^{2} - 3y + 1 = 0$