If #8sqrt(4z^2 - 43) = 40#, what is the value of #z#?

2 Answers
Aug 27, 2016

Assumption: the question is: #" "8sqrt(4z^2)-43=40#

#z=+83/16 ->5 3/16#

Explanation:

#color(brown)("The objective is to manipulate the equation such that you have a")##color(brown)("single z. This is to be on one side of the equals sign and everything")# #color(brown)("else on the other side.")#

This is done in stages. First you have all the terms with z on the LHS of = and all the terms without z on the other side.
Then you manipulate the LHS side until the only thing left is the single z.

#color(blue)("Step 1 - Isolate "sqrt(4z^2)#

Add 43 to both sides giving

#8sqrt(4z^2)+0=40+43#

Divide both sides by 8. Same as multiply by #1/8# to get rif og the 8 from #8sqrt(4z^2)#

#8/8xxsqrt(4z^2)=83/8#

But #8/8=1#

#sqrt(4z^2)=83/8#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Step 2 - Isolate z")#

But #4 = 2^2 -> 4z^2 = 2^2z^2#

Write as:#" "sqrt(2^2z^2)=83/8#

Taking the root

#=>+-2z=83/8#

divide both side by 2 to get rid of the 2 from #2z#

#+-z=83/16-> 5 3/16#

After testing #z=+83/16#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Check: Consider the left hand side only for #z=+83/16#

#8sqrt(4z^2)-43#

#8xx(2z)-43#

#8xx2xx83/16-43#

But #2xx8=16#

#cancel(16)xx83/(cancel(16))-43#

#83-43#

#40#

Thus LHS=RHS so

#color(red)(" The equation will not works for "z=-83/16)#

Aug 27, 2016

#z = +-sqrt(17)#

Explanation:

#8sqrt(4z^2 - 43) = 40#

Isolate the square-root.

#sqrt(4z^2 - 43) = 5#

#(sqrt(4z^2 - 43))^2 = 5^2#

#4z^2 - 43 = 25#

#4z^2 = 25 + 43#

#4z^2 = 68#

#z^2 = 17#

#z = +-sqrt(17#

Checking in the original equation, both solutions work.