How do you solve #x^ { 2} - 16x - 5= - 21#?

1 Answer
Dan Kemp
Jun 11, 2017

#x = 8 - 4sqrt3 or 8 + 4sqrt3#

Explanation:

#x^2-16x - 5 = -21#

#x^2-16x = -21 + 5#

#x^2-16x = -16#

We can use the quadratic formula to solve.

#x^2-16x color(red)( +16)= color(red)cancel(color(black)(-16) color(red)( +16)#

#x^2-16x + 16 = 0#

#ax^2 + bx + c = 0#

#a = 1#
#b = -16#
#c = 16#

# x = (-b +- sqrt(b^2-4ac)) / (2a) #

# x = (color(red)(cancel(color(black)( - -))) 16 +- sqrt((-16)^2-4 color(red)(cancel(color(black)(xx 1))) xx 16)) / (2 color(red)(cancel(color(black)(xx 1) #

# x = (16 +- sqrt((-16)^2 - 4 xx 16)) / 2 #

# x = (16 +- sqrt(256 - 4 xx 16)) / 2 #

# x = (16 +- sqrt(256 - 64)) / 2 #

# x = (16 +- sqrt192) / 2 #

# x = (16 +- 8sqrt3) / 2 = 16/2 +- (8sqrt3) / 2 = 8 +- 4sqrt3#

#color(blue)(x = 8 - 4sqrt3 or 8 + 4sqrt3#

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