# How do you solve #4sqrt(x+3)=x#?

##### 1 Answer

May 18, 2015

Answer: x=4 and x=12

First you want to square both sides to get rid of the square root aka the radical. Once you square both sides you should get: 16(x+3)=x^2. Now multiple the 16 by the (x+3) and you will get 16x+48.

Your equation should now look like this:

16x+48=x^2

Now you want to get all of your numbers on side, so subtract the 16x+48 from the left side and bring it to the right. Your equation will now look like this:

0=x^2-16x-48

Now you must factor to solve for x!

- 0=x^2-16x-48
- 0= (x-4) (x-12)
- 0= (x-4) and 0=(x-12)

Now solve for x by adding 4 and 12 to zero. - x=4, and x=12.