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 $x=1, b=-4, c=5$

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

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

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

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

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