How do you evaluate #3 + sqrt(x+7) - sqrt(3x)# when x = 9?

3 Answers
NJ
May 10, 2018

#=>7-3sqrt(3)#

Explanation:

#3+ sqrt(x+7) - sqrt(3x)#

When #x = 9#, we get

#3+sqrt(9+7) - sqrt(3*9)#

#=3+sqrt(16) - sqrt(3*3^2)#

#=3+4 - 3sqrt(3)#

#=7-3sqrt(3)#

Karla · Stefan V.
May 10, 2018

# 7-3 * (3)^(1/2)#

Explanation:

Put #9# in place of #x#

#3+(9+7)^(1/2) -(3*9)^(1/2)#

It will be

#3+16^(1/2) -27^(1/2)#

As

#16^(1/2)=4 #

Thus

#3+4-27^(1/2)#

It will be

#7-27^(1/2)#

As #27^(1/2)# can be simplified to #(3*9)^(1/2)# and as #3*9=27#, it will be the same as

#3^(1/2) * 9^(1/2)#

Since

#9^(1/2) =3 #

It will be

#3^(1/2) * 3#

Coming back to

#7-27^(1/2)#

As

#27^(1/2) = 3 * 3^(1/2)#

So it will be

#7-3 * 3^(1/2)#

Jacobi J.
May 10, 2018

#=>7-3sqrt3#

Explanation:

Plugging in #9# for #x#, we get

#3+sqrt(9+7)-sqrt(3*9)#

which simplifies to

#3+sqrt(16)-sqrt(27)#

#sqrt16# evaluates to #4#. Thus we have

#3+4-sqrt(27)#

#=>7-sqrt(27)#

We can rewrite #sqrt27# as #sqrt(9)*sqrt3#. We get

#7-(sqrt9*sqrt3)#

#=>7-3sqrt3#

Hope this helps!

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