How do you solve #sqrt(9u-4)=sqrt(7u-20)#?

Question

1849 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-24T14:18:45

See a solution process below:

First, square both sides of the equation to eliminate the radicals while keeping the equation balanced:

#(sqrt(9u - 4))^2 = (sqrt(7u - 20))^2#

#9u - 4 = 7u - 20#

Next, add #color(red)(4)# and subtract #color(blue)(7u)# from each side of the equation to isolate the #u# term while keeping the equation balanced:

#9u - color(blue)(7u) - 4 + color(red)(4) = 7u - color(blue)(7u) - 20 + color(red)(4)#

#(9 - color(blue)(7))u - 0 = 0 - 16#

#2u = -16#

Now, divide each side of the equation by #color(red)(2)# to solve for #u# while keeping the equation balanced:

#(2u)/color(red)(2) = -16/color(red)(2)#

#(color(red)(cancel(color(black)(2)))u)/cancel(color(red)(2)) = -8#

#u = -8#

Answer 2 · 2018-01-24T14:30:18

See below.

#sqrt(9u-4)=sqrt(7u-20)#

Start by squaring both sides. This will remove the radicals:

#(sqrt(9u-4))^2=(sqrt(7u-20))^2#

#9u-4=7u-20#

Collect all terms containing the variable on one side of the equation, and all constant terms on the opposite side:

#9u-7u=-20+4#

Simplify by adding like terms:

#2u=-16#

We need to get a #u# on its own.

Divide both sides by #2#:

#(2u)/2=(-16)/2->(cancel(2)u)/cancel(2)=(-cancel(16)8)/cancel(2)->color(blue)(u=-8)#