How do you solve ${(x - 3)}^{2} + 25 = 0$ $(x-3)^2+25=0$ ?

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Answer 1 · 2014-11-14T20:40:22

If you are looking for real solutions, then there are none.

If you are looking for complex solutions, then

${(x - 3)}^{2} + 25 = 0$ $(x-3)^2+25=0$

by subtracting $25$ $25$ ,

$\Rightarrow {(x - 3)}^{2} = - 25$ $=> (x-3)^2=-25$

by taking the square-root,

$\Rightarrow x - 3 = \pm \sqrt{- 25} = \pm 5 i$ $=>x-3=pmsqrt{-25}=pm5i$

by adding $3$ $3$ ,

$\Rightarrow x = 3 \pm 5 i$ $=> x=3 pm 5i$

I hope that this was helpful.

How do you solve (x−3)2+25=0(x-3)^2+25=0?