How do you simplify #sqrt12-sqrt3#?

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Answer 1 · 2018-03-10T18:06:30

See a solution process below:

First rewrite the radical on the left as:

#sqrt(4 * 3) - sqrt(3)#

Now, use this rule of radicals to simplify the radical on the left:

#sqrt(color(red)(a) * color(blue)(b)) = sqrt(color(red)(a)) * sqrt(color(blue)(b))#

#sqrt(color(red)(4) * color(blue)(3)) - sqrt(3) =>#

#sqrt(color(red)(4))sqrt(color(blue)(3)) - sqrt(3) =>#

#2sqrt(color(blue)(3)) - sqrt(3)#

We can now factor out the common term:

#2sqrt(color(blue)(3)) - sqrt(color(blue)(3)) =>#

#2sqrt(color(blue)(3)) - 1sqrt(color(blue)(3)) =>#

#(2 - 1)sqrt(color(blue)(3)) =>#

#1sqrt(color(blue)(3)) =>#

#sqrt(color(blue)(3))#

Answer 2 · 2018-03-10T18:08:07

#sqrt(3)#

Simplify #sqrt(12)#

#sqrt(4)sqrt(3)#

And further, simplify to get

#2sqrt(3)#

Subtract #sqrt(3)# from #2sqrt(3)#

#sqrt(3)#