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5 votes
What's the possible values of z when t = 81

( be it real or complex number )


{z}^(2) = √(t)
​​
So annoying ~

User Mstuercke
by
8.4k points

1 Answer

5 votes

Answer:


  • \boxed{\mathrm{z=\pm 3}}

Explanation:


\mathrm{z^2=√(t), \; when, \; t=81}

Substitute t with 81:-


\mathrm{z^2=√(81)}

Since 9*9 =81, square of 81 is 9:-


\mathrm{z^2=9}

Take the square root of both sides.


\mathrm{z=\pm √(9) }

and since 3*3 = 9 the square root of 9 will be 3.:-


\mathrm{z=\pm 3}

_____________________

Hope this helps!

User Alexey Borovikov
by
8.2k points

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