How do you solve $- 2 x^{2} + 5 x + \frac{1}{8}$ $-2x^2 + 5x + 1/8$ by completing the square?

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Answer 1 · 2015-05-01T03:00:46

Change side and divide by a:

$f (x) = x^{2} - \frac{5}{2} x - \frac{1}{16} = 0$ $f(x) = x^2 - 5/2 x - 1/16 = 0$

$f (x) = x^{2} - \frac{5}{2} x + \frac{25}{16} - \frac{25}{16} - \frac{1}{16} = 0$ $f(x) = x^2 - 5/2 x + 25/16 - 25/16 - 1/16 = 0$

${(x - \frac{5}{4})}^{2} = \frac{25}{16} + \frac{1}{16} = \frac{26}{16}$ $( x - 5/4)^2 = 25/16 + 1/16 = 26/16$

$(x - \frac{5}{4}) = \frac{s q r 26}{4} - \to x = \frac{5}{4} + \frac{s q r 26}{4}$ $(x - 5/4) = (sqr26)/4 --> x = 5/4 + (sqr26)/4$

$(x - \frac{5}{4}) = \frac{- s q r 26}{4} - \to x = \frac{5}{4} - \frac{s q r 26}{4}$ $(x - 5/4) = (-sqr26)/4 --> x = 5/4 - (sqr26)/4$

How do you solve −2x2+5x+18-2x^2 + 5x + 1/8 by completing the square?