spacewater
Junior Member
- Joined
- Jul 10, 2009
- Messages
- 67
problem
\(\displaystyle \sqrt {2x+6} - \sqrt{x+4} = 1\)
I got this part .. to free \(\displaystyle \sqrt{2x+6}\) by itself and square it, but I dont get how the right side of equation comes out to
\(\displaystyle \sqrt{2x+6} = 1 + \sqrt{x+4}\)
\(\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)\) <------ I dont understand how this came out to be like this shouldnt it be \(\displaystyle 2x + 6 = 1 + x + 4\) since we're squaring both side?
Can someone who isn't too busy elaborate this problem to me please?
\(\displaystyle \sqrt {2x+6} - \sqrt{x+4} = 1\)
I got this part .. to free \(\displaystyle \sqrt{2x+6}\) by itself and square it, but I dont get how the right side of equation comes out to
\(\displaystyle \sqrt{2x+6} = 1 + \sqrt{x+4}\)
\(\displaystyle 2x+6 = 1+ 2\sqrt{x+4} + (x+4)\) <------ I dont understand how this came out to be like this shouldnt it be \(\displaystyle 2x + 6 = 1 + x + 4\) since we're squaring both side?
Can someone who isn't too busy elaborate this problem to me please?