The minimum value of 2 $x^{2} + x - 1$ is

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Solution

$\begin{matrix} 2 x^{2} + x - 1 2 [x^{2} + \frac{x}{2}] - 1 = 2 [{(x + \frac{1}{4})}^{2} - \frac{1}{16}] - 1 = 2 {(x + \frac{1}{4})}^{2} - \frac{1}{8} - 1 = 2 {(x + \frac{1}{4})}^{2} - \frac{9}{8} m i n i m u m v a l u e = \frac{- 9}{8} a t x = - \frac{1}{4} \end{matrix}$

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