What is the square root of -8?

Question

Kennedy Adams · Accepted Answer

undefined The square root of -8 is not a real number because the square root of a negative number is not defined within the set of real numbers. However, in the realm of complex numbers, the square root of -8 can be expressed as 2i√2 or -2i√2, where i represents the imaginary unit (√(-1)).

Ethan Adkins · Answer

#pm2isqrt2#

We have the following:

#sqrt(-8)#

We can rewrite this as

#color(steelblue)(sqrt8)*color(purple)(sqrt(-1))#

Recall that #i=sqrt(-1)#

We can further break this down as

#color(steelblue)(sqrt(4)*sqrt2)*color(purple)i#, or

#pm2isqrt2#

Hope this helps!

Eva Abramson · Answer

#\sqrt{-8}=\pm 2i\sqrt2# The square root of #-8# is given #\sqrt{-8}# #=(-8)^{1/2}# #=(8(\cos\pi+i\sin\pi))^{1/2}# #=8^{1/2}(\cos\pi+i\sin\pi)^{1/2}# #=2\sqrt2(\cos(2k\pi+\pi)+i\sin(2k\pi+\pi))^{1/2}# #=2\sqrt2(\cos((2k+1)\pi)+i\sin((2k+1)\pi))^{1/2}# #=2\sqrt2(\cos({(2k+1)\pi}/2)+i\sin({(2k+1)\pi}/2))# Where, #k=0, 1# Now, setting #k=0# & #k=1#, we get two values as follows #\sqrt{-8}=2\sqrt2(\cos({(2(0)+1)\pi}/2)+i\sin({(2(0)+1)\pi}/2))# #\sqrt{-8}=2i\sqrt2# & #\sqrt{-8}=2\sqrt2(\cos({(2(1)+1)\pi}/2)+i\sin({(2(1)+1)\pi}/2))# #\sqrt{-8}=-2i\sqrt2# # herefore \sqrt{-8}=\pm 2i\sqrt2#

Savannah Anderson · Answer

#color(green)(sqrt(-8)=2sqrt2i or sqrt(-8)=-2sqrt2i#

Let ,

#x=sqrt(-8)...tox!inRR, but , x in CC#

Squaring both sides

#x^2=-8=8(-1)#

#:.x^2=8i^2 to[because i^2=-1]#

#:.x^2-8i^2=0#

#:.x^2-(2sqrt2 i)^2=0#

#:.(x-2sqrt2i)(x+2sqrt2i)=0to[color(blue)(because a^2- b^2#=#color(blue)((a-b)(a+b))]#

#:.x-2sqrt2i=0 or x+2sqrt2i=0#

#:.x=2sqrt2i or x=-2sqrt2i#

#:.color(green)(sqrt(-8)=2sqrt2i or sqrt(-8)=-2sqrt2i# ...................................................................................

Note: If #x=sqrt8=sqrt(4 xx2)=sqrt((2sqrt2)^2)#

#:.color(red)(x=sqrt8=+-2sqrt2 inRR#