How do you solve #46- 4x = 4x ^ { 2}#?

2 Answers
salamat
Apr 18, 2017

#x = (-1 +- sqrt 47)/2#

Explanation:

#46 - 4x = 4x^2#
#4x^2 + 4x - 46 =0#

reduce the equation by divide with #2#
#2x^2 + 2x - 23 = 0#

#2(x^2 + x ) -23 =0#

#2(x + 1/2)^2 -2(1/2)^2 -23 = 0#

#2(x + 1/2)^2 - 47/2 = 0#

#2(x + 1/2)^2 = 47/2#

#(x + 1/2)^2 = 47/4#

#x + 1/2 = +-sqrt(47/4) = +-sqrt47/2#

#x = -1/2 +-sqrt47/2 = (-1 +- sqrt 47)/2#

Catherine T. Mohs
Apr 18, 2017

#x=-1/2+-sqrt(47)/2#

Explanation:

#46-4x=4x^2#
#4x^2+4x-46=0#
#x=(-b+-sqrt(b^2-4ac))/(2a)#
#x=(-4+-sqrt(4^2-4*4*-46))/(2*4)#
#x=(-4+-sqrt(16+736))/(8)#
#x=(-4+-sqrt(752))/(8)#
#x=-1/2+-sqrt(47)/2#

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