How do you complete the square to solve $- 2 x^{2} - 7 x + 4 = 0$ $-2x^2 - 7x + 4 = 0$ ?

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How do you complete the square to solve $- 2 x^{2} - 7 x + 4 = 0$ $-2x^2 - 7x + 4 = 0$ ?

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Answer 1 · 2015-06-01T16:53:30

$f (x) = - 2 (x^{2} + 7 \frac{x}{2} + \frac{49}{16}) - \frac{49}{16} - 2) = 0$ $f(x) = -2(x^2 + 7x/2 + 49/16) - 49/16 - 2)= 0$

$- 2 {(x + \frac{7}{4})}^{2} + \frac{98}{16} + 4$ $-2(x + 7/4)^2 + 98/16 + 4$

$- 2 {(x + \frac{7}{4})}^{2} = - \frac{98}{16} - 4 = - \frac{162}{16}$ $-2(x + 7/4)^2 = -98/16 - 4 = -162/16$

${(x + \frac{7}{4})}^{2} = \frac{81}{16} \to (x + \frac{7}{4}) \pm \frac{9}{4}$ $(x + 7/4)^2 = 81/16 -> (x + 7/4) +- 9/4$

$x = - \frac{7}{4} \pm \frac{9}{4}$ $x = -7/4 +- 9/4$ -> x = -4 and x = 1/2

NOTE. This method of completing the squares is unadvised when b is an odd number and a is negative. Completing the squares in this case is very challenging and easy to make errors/mistakes.
Solving by the new Transforming method (Google, Yahoo Search) is a lot simpler.
y = -(2x^2 + 7x - 4) = 0 (1)
y' = -(x^2 + 7x - 8) (2) -> a.c = -8 --> factor: (-1, 8) -> sum = b
Two real roots of (2): 1 and -8
Two real roots of (1): 1/2 and -4

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How do you complete the square to solve $- 2 x^{2} - 7 x + 4 = 0$ $-2x^2 - 7x + 4 = 0$ ?

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