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

Question

Kennedy Adams · Accepted Answer

#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#

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#

Eli Assouline · Answer

undefined To solve the equation $ (x^2 - 4x)^2 - 25 = 0 $, follow these steps:

1. Expand $ (x^2 - 4x)^2 $ to get $ x^4 - 8x^3 + 16x^2 $.
2. Substitute the expanded expression into the equation to get $ x^4 - 8x^3 + 16x^2 - 25 = 0 $.
3. Rearrange the terms to get $ x^4 - 8x^3 + 16x^2 - 25 = 0 $.
4. Factor the quadratic expression $ x^4 - 8x^3 + 16x^2 - 25 $ to get $ (x^2 - 4x - 5)(x^2 - 4x + 5) = 0 $.
5. Set each factor equal to zero and solve for $ x $:
   $ x^2 - 4x - 5 = 0 $ and $ x^2 - 4x + 5 = 0 $.
6. Solve each quadratic equation using the quadratic formula or factoring method to find the values of $ x $.

Abigail Allen · Answer

undefined To solve the equation (x^2 - 4x)^2 - 25 = 0, follow these steps:

1. Expand the square: (x^2 - 4x)^2 = (x^2 - 4x)(x^2 - 4x) = x^4 - 8x^3 + 16x^2.
2. Substitute the expanded expression into the original equation: x^4 - 8x^3 + 16x^2 - 25 = 0.
3. Rearrange the equation: x^4 - 8x^3 + 16x^2 - 25 = 0 becomes x^4 - 8x^3 + 16x^2 = 25.
4. Move 25 to the other side of the equation: x^4 - 8x^3 + 16x^2 - 25 = 0 becomes x^4 - 8x^3 + 16x^2 - 25 = 0.
5. Factor the equation if possible or solve using numerical methods.
   
There isn't a simple factorization for this quartic equation, so you'd typically solve it using numerical methods like Newton's method or using a graphing calculator or software to find the approximate solutions.