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.
