How do you solve #root3(6u-5)+2=-3#?

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Answer 1 · 2017-01-15T18:13:40

Move the 2 to the right.
Cube both sides.
Then solve for u.

Given: #root(3)(6u - 5) + 2 = -3#

Subtract 2 from both sides:

#root(3)(6u - 5) = -5#

Cube both sides:

#6u - 5 = -125#

Add 5 to both sides:

#6u = -120#

Divide both sides by 6:

#u = -20#

check:

#root(3)(6(-20) - 5) + 2 = -3#

#root(3)(-120 - 5) + 2 = -3#

#root(3)(-125) + 2 = -3#

#-5 + 2 = -3#

#-3 = -3#

This checks.