How do you solve #-48z ^ { 2} = 3#?

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Answer 1 · 2017-09-02T17:54:25

See a solution process below:

First, divide each side of the equation by #color(red)(-48)# to isolate #z^2# while keeping the equation balanced:

#(-48z^2)/color(red)(-48) = 3/color(red)(-48)#

#(color(red)(cancel(color(black)(-48)))z^2)/cancel(color(red)(-48)) = -1/16#

#z^2 = -1/16#

Next take the square root of each side of the equation to solve for #z# while keeping the equation balanced:

#sqrt(z^2) = +-sqrt(-1/16)#

#z = +-sqrt(-1/16)#

#z = +-sqrt(1/16 xx -1)#

#z = +-sqrt(1/16)sqrt(-1)#

#z = +-1/4sqrt(-1)#

#z = +-sqrt(-1)/4#

However, we cannot take the square root of a negative number therefore there are no Real solutions to this problem.

Answer 2 · 2017-09-02T18:02:19

#z = +-1/4i#

Given:

#-48z^2 = 3#

Divide both sides by #-48# to get:

#z^2 = -1/16#

The square of any real number is non-negative, so this has no real solutions.

It does have two complex solutions, involving the imaginary unit, which satisfies #i^2=-1#.

Then we find:

#(1/4i)^2 = (1/4)^2 i^2 = 1/16 (-1) = -1/16#

So: #z = 1/4i# is a solution.

We also find:

#(-1/4i)^2 = (-1/14)^2 i^2 = 1/16 (-1) = -1/16#

So: #z = -1/4i# is also a solution.