How do you solve #6= \sqrt { - 5- 3u } + 2#?

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Answer 1 · 2017-05-29T09:32:25

u = - 7

subtract (-2) from both sides, we get

#6-2=sqrt(-5-3u) +2-2#

#rArr 4= sqrt(-5-3u)# [ Note: squaring both sides]

#rArr 4^2=[sqrt(-5-3u)]^2#

#rArr 16 = -5-3u#

#rArr 3u=-5-16 or -21#

#rArr u -21/3 or -7#

Answer 2 · 2017-05-29T09:36:10

#u = -7#

#6 = sqrt(-5 - 3 u) + 2#

We can use some simple algebra to work this out.

#6 = sqrt(-5 - 3 u) + 2#

#6 -2= sqrt(-5 - 3 u)#

#4= sqrt(-5 - 3 u)#

#4^2 = -5 - 3 u#

#16 = -5 - 3 u#

#16 + 5= - 3 u#

#21= - 3 u#

#u = 21 ÷ -3 #

#color(blue)(u = -7#

We can now substitute #u# for #-7# to prove that we are correct.

#6 = sqrt(-5 - 3 u) + 2#

#6 = sqrt(-5 - 3 xx 7) + 2#

#6 = sqrt16 + 2#

#6 = 4 + 2#

#6 = 6#