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

Jun 5, 2018

$x = 5 \mathmr{and} x = - 1$ and if imaginary solutions are allowed, $x = 2 + i , x = 2 - i$

#### Explanation:

${\left({x}^{2} - 4 x\right)}^{2} = 25$

Square root both sides

${x}^{2} - 4 x = \setminus \pm 5$

${x}^{2} - 4 x - 5 = 0 \mathmr{and} {x}^{2} - 4 x + 5 = 0$

$\left(x - 5\right) \left(x + 1\right) = 0$ but ${x}^{2} - 4 x + 5 = 0$ has no real solutions

$x = 5 \mathmr{and} x = - 1$

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

$x = \frac{- b \setminus \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

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

$x = \frac{4 \setminus \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(1\right) \left(5\right)}}{2 \left(1\right)}$

$x = \frac{4 \setminus \pm \sqrt{16 - 20}}{2}$

$x = \frac{4 \setminus \pm \sqrt{- 4}}{2}$

$x = \frac{4 \setminus \pm 2 i}{2} \implies x = 2 + i , x = 2 - i$