How do you solve #(x^2-4x)^(2)-25=0#?

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Answer 1 · 2018-06-05T20:17:33

#x=5 or x=-1# and if imaginary solutions are allowed, #x=2+i, x=2-i#

add 25 to both sides

#(x^2-4x)^2=25#

Square root both sides

#x^2-4x=\pm5#

#x^2-4x-5=0 or x^2-4x+5=0#

#(x-5)(x+1)=0# but #x^2-4x+5=0# has no real solutions

#x=5 or x=-1#

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If imaginary solutions are allowed, then the solutions to #x^2-4x+5=0# are (using the quadratic formula):

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

with #x=1, b=-4, c=5#

# x = (4 \pm sqrt((-4)^2-4(1)(5))) / (2(1)) #

# x = (4 \pm sqrt(16-20)) / 2 #

# x = (4 \pm sqrt(-4)) / 2 #

# x = (4 \pm 2i) / 2 =>x=2+i, x=2-i#