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

1 Answer
Nghi N.
Jun 1, 2015

#f(x) = -2(x^2 + 7x/2 + 49/16) - 49/16 - 2)= 0#

#-2(x + 7/4)^2 + 98/16 + 4 #

#-2(x + 7/4)^2 = -98/16 - 4 = -162/16 #

#(x + 7/4)^2 = 81/16 -> (x + 7/4) +- 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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