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Question Number 63539 by REHEMAABDDUL last updated on 05/Jul/19

$$\mathrm{The}\mathrm{minimum}\mathrm{value}\mathrm{of}2x^2-3x+2\mathrm{is}\mathrm{\_}\mathrm{\_}\mathrm{\_}.$$

Commented by Prithwish sen last updated on 05/Jul/19

$$=2{(\mathrm{x}-\frac{3}{4})}^2+\frac{7}{8}$$$$\because 2{(\mathrm{x}-\frac{3}{4})}^2⩾0$$$$\therefore \mathrm{the}\mathrm{minimum}\mathrm{value}=\frac{7}{8}$$

Commented by meme last updated on 05/Jul/19

$$a=2>0$$$$theminimumisx=\frac{3}{4}$$

Answered by Rio Michael last updated on 05/Jul/19

$$minvalue=\frac{4ac-b^2}{4a}$$$$minval=\frac{4\left(2\right)\left(2\right)-{(-3)}^2}{4\left(2\right)}$$$$=\frac{7}{8}$$

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